· . t f (x) e e ng f (10) e coe h 49598666989151226098104244512918, n f (x) e 2 e...
TRANSCRIPT
Regions
containing
roots
ofpolynom
ials
MichaelFilaseta
University
ofSouth
Carolina
Regions
containing
roots
ofpolynom
ials
MichaelFilaseta
University
ofSouth
Carolina
Regions
containing
roots
ofpolynom
ials
MichaelFilaseta
University
ofSouth
Carolina
Sim
ple
ClassicalR
esult.
Let
f(x)
be
monic
in
C[x]with
deg
f�
1and
|f(0)|
�1.
Then
f(x)
has
aroot
on
or
outside
the
unitcircle
D=
{z
2C
:|z|
=1}
.
Corollary.
Every
non-constantm
onic
polynom
ial
f(x)
2Z[x]has
arooton
or
outside
D=
{z
2C
:|z|
=1}
.
Theorem
(K
ronecker).
If
f(x)
2Z[x]
is
monic,
irre-
ducible,and
has
allits
roots
on
D=
{z
2C
:|z|
=1}
,
then
f(x)
is
acyclotom
ic
polynom
ial.
TypicalD
iscussion:
Mahler
measure
Lehm
er’s
polynom
ial
Etc.
Sim
ple
ClassicalR
esult.
Let
f(x)
be
monic
in
C[x]with
deg
f�
1and
|f(0)|
�1.
Then
f(x)
has
aroot
on
or
outside
the
unitcircle
C=
{z
2C
:|z|
=1}
.
Corollary.
Every
non-constantm
onic
polynom
ial
f(x)
2Z[x]has
arooton
or
outside
D=
{z
2C
:|z|
=1}
.
Theorem
(K
ronecker).
If
f(x)
2Z[x]
is
monic,
irre-
ducible,and
has
allits
roots
on
D=
{z
2C
:|z|
=1}
,
then
f(x)
is
acyclotom
ic
polynom
ial.
TypicalD
iscussion:
Mahler
measure
Lehm
er’s
polynom
ial
Etc.
Sim
ple
ClassicalR
esult.
Let
f(x)
be
monic
in
C[x]with
deg
f�
1and
|f(0)|
�1.
Then
f(x)
has
aroot
on
or
outside
the
unitcircle
C=
{z
2C
:|z|
=1}
.
Corollary.
Every
non-constantm
onic
polynom
ial
f(x)
2Z[x]has
arooton
or
outside
C=
{z
2C
:|z|
=1}
.
Theorem
(K
ronecker).
If
f(x)
2Z[x]
is
monic,
irre-
ducible,
and
has
allits
roots
on
C=
{z
2C
:|z|
=1}
,
then
f(x)
is
acyclotom
ic
polynom
ial.
TypicalD
iscussion:
Mahler
measure
Lehm
er’s
polynom
ial
Etc.
Sim
ple
ClassicalR
esult.
Let
f(x)
be
monic
in
C[x]with
deg
f�
1and
|f(0)|
�1.
Then
f(x)
has
aroot
on
or
outside
the
unitcircle
C=
{z
2C
:|z|
=1}
.
Corollary.
Every
non-constant
monic
f(x)
2Z[x]
with
f(0)
6=0
has
arooton
or
outside
C=
{z
2C
:|z|
=1}
.
Theorem
(K
ronecker).
If
f(x)
2Z[x]
is
monic,
irre-
ducible,
and
has
allits
roots
on
C=
{z
2C
:|z|
=1}
,
then
f(x)
is
acyclotom
ic
polynom
ial.
TypicalD
iscussion:
Mahler
measure
Lehm
er’s
polynom
ial
Etc.
Can
we
say
som
ething
di↵erent
about
f(x)
2Z[x]?
RCom
ment.Every
non-constantm
onic
polynom
ialf(x)
2Z[x]has
arooton
or
outside
the
region
R.
Sim
ple
ClassicalR
esult.
Let
f(x)
be
monic
in
C[x]with
deg
f�
1and
|f(0)|
�1.
Then
f(x)
has
aroot
on
or
outside
the
unitcircle
C=
{z
2C
:|z|
=1}
.
Corollary.
Every
non-constantm
onic
polynom
ial
f(x)
2Z[x]has
arooton
or
outside
D=
{z
2C
:|z|
=1}
.
Theorem
(K
ronecker).
If
f(x)
2Z[x]
is
monic,
irre-
ducible,and
has
allits
roots
on
D=
{z
2C
:|z|
=1}
,
then
f(x)
is
acyclotom
ic
polynom
ial.
TypicalD
iscussion:
Mahler
measure
Lehm
er’s
polynom
ial
Etc.
Sim
ple
ClassicalR
esult.
Let
f(x)
be
monic
in
C[x]with
deg
f�
1and
|f(0)|
�1.
Then
f(x)
has
aroot
on
or
outside
the
unitcircle
C=
{z
2C
:|z|
=1}
.
Corollary.
Every
non-constant
monic
f(x)
2Z[x]
with
f(0)
6=0
has
arooton
or
outside
C=
{z
2C
:|z|
=1}
.
Theorem
(K
ronecker).
If
f(x)
2Z[x]
is
monic,
irre-
ducible,
and
has
allits
roots
on
C=
{z
2C
:|z|
=1}
,
then
f(x)
is
acyclotom
ic
polynom
ial.
TypicalD
iscussion:
Mahler
measure
Lehm
er’s
polynom
ial
Etc.
Suppose
f(x)
2Z[x]and
|f(0)|
=1.
Know
:f(x)
has
aroot
in
D=
{z
2C
:|z|
1}.
Pictures
ofother
regions
thatm
ustcontain
arootoff(x).
Where
did
these
regions
com
efrom
?
Why
are
they
ofinterest
to
me?
What
are
or
are
not
other
possible
regions
ofinterest?
Supposef(x)2Z[x]and|f(0)|=1.
Know:f(x)hasarootinD={z2C:|z|1}.
Picturesofotherregionsthatmustcontainarootoff(x).
Wheredidtheseregionscomefrom?
Whyaretheyofinteresttome?
Whatareorarenototherpossibleregionsofinterest?
arootinhere
=)
Suppose
f(x)
2Z[x]and
|f(0)|
=1.
Know
:f(x)
has
aroot
in
D=
{z
2C
:|z|
1}.
Pictures
ofother
regions
thatm
ustcontain
arootoff(x).
Where
did
these
regions
com
efrom
?
Why
are
they
ofinterest
to
me?
What
are
or
are
not
other
possible
regions
ofinterest?
aroot
out
here
f(x)
monic
=)
Suppose
f(x)
2Z[x]and
|f(0)|
=1.
Know
:f(x)
has
aroot
in
D=
{z
2C
:|z|
1}.
Pictures
ofother
regions
thatm
ustcontain
arootoff(x).
Where
did
these
regions
com
efrom
?
Why
are
they
ofinterest
to
me?
What
are
or
are
not
other
possible
regions
ofinterest?
aroot
in
here
f(x)
monic
=)
Suppose
f(x)
2Z[x]and
|f(0)|
=1.
Know
:f(x)
has
aroot
in
D=
{z
2C
:|z|
1}.
Pictures
ofother
regions
thatm
ustcontain
arootoff(x).
Where
did
these
regions
com
efrom
?
Why
are
they
ofinterest
to
me?
What
are
or
are
not
other
possible
regions
ofinterest?
Suppose
f(x)
2Z[x]is
not
constant
and
|f(0)|
=1.
Know
:f(x)
has
aroot
in
D=
{z
2C
:|z|
1}.
Pictures
ofother
regions
thatm
ustcontain
arootoff(x).
Where
did
these
regions
com
efrom
?
Why
are
they
ofinterest
to
me?
What
are
or
are
not
other
possible
regions
ofinterest?
aroot
out
here
f(x)
monic
=)
Sim
ple
ClassicalR
esult.
Let
f(x)
be
monic
in
C[x]with
deg
f�
1and
|f(0)|
�1.
Then
f(x)
has
aroot
on
or
outside
the
unitcircle
C=
{z
2C
:|z|
=1}
.
Corollary.
Every
non-constant
monic
f(x)
2Z[x]
with
f(0)
6=0
has
arooton
or
outside
C=
{z
2C
:|z|
=1}
.
Theorem
(K
ronecker).
If
f(x)
2Z[x]
is
monic,
irre-
ducible,
and
has
allits
roots
on
C=
{z
2C
:|z|
=1}
,
then
f(x)
is
acyclotom
ic
polynom
ial.
TypicalD
iscussion:
Mahler
measure
Lehm
er’s
polynom
ial
Etc.
Suppose
f(x)
2Z[x]and
|f(0)|
=1.
Know
:f(x)
has
aroot
in
D=
{z
2C
:|z|
1}.
Pictures
ofother
regions
thatm
ustcontain
arootoff(x).
Where
did
these
regions
com
efrom
?
Why
are
they
ofinterest
to
me?
What
are
or
are
not
other
possible
regions
ofinterest?
Supposef(x)2Z[x]and|f(0)|=1.
Know:f(x)hasarootinD={z2C:|z|1}.
Picturesofotherregionsthatmustcontainarootoff(x).
Wheredidtheseregionscomefrom?
Whyaretheyofinteresttome?
Whatareorarenototherpossibleregionsofinterest?
arootinhere
=)
Suppose
f(x)
2Z[x]and
|f(0)|
=1.
Know
:f(x)
has
aroot
in
D=
{z
2C
:|z|
1}.
Pictures
ofother
regions
thatm
ustcontain
arootoff(x).
Where
did
these
regions
com
efrom
?
Why
are
they
ofinterest
to
me?
What
are
or
are
not
other
possible
regions
ofinterest?
aroot
out
here
f(x)
monic
=)
Suppose
f(x)
2Z[x]and
|f(0)|
=1.
Know
:f(x)
has
aroot
in
D=
{z
2C
:|z|
1}.
Pictures
ofother
regions
thatm
ustcontain
arootoff(x).
Where
did
these
regions
com
efrom
?
Why
are
they
ofinterest
to
me?
What
are
or
are
not
other
possible
regions
ofinterest?
aroot
in
here
f(x)
monic
=)
Suppose
f(x)
2Z[x]and
|f(0)|
=1.
Know
:f(x)
has
aroot
in
D=
{z
2C
:|z|
1}.
Pictures
ofother
regions
thatm
ustcontain
arootoff(x).
Where
did
these
regions
com
efrom
?
Why
are
they
ofinterest
to
me?
What
are
or
are
not
other
possible
regions
ofinterest?
Suppose
f(x)
2Z[x]is
not
constant
and
|f(0)|
=1.
Know
:f(x)
has
aroot
in
D=
{z
2C
:|z|
1}.
Pictures
ofother
regions
thatm
ustcontain
arootoff(x).
Where
did
these
regions
com
efrom
?
Why
are
they
ofinterest
to
me?
What
are
or
are
not
other
possible
regions
ofinterest?
aroot
out
here
f(x)
monic
=)
Sim
ple
ClassicalR
esult.
Let
f(x)
be
monic
in
C[x]with
deg
f�
1and
|f(0)|
�1.
Then
f(x)
has
aroot
on
or
outside
the
unitcircle
C=
{z
2C
:|z|
=1}
.
Corollary.
Every
non-constant
monic
f(x)
2Z[x]
with
f(0)
6=0
has
arooton
or
outside
C=
{z
2C
:|z|
=1}
.
Theorem
(K
ronecker).
If
f(x)
2Z[x]
is
monic,
irre-
ducible,
and
has
allits
roots
on
C=
{z
2C
:|z|
=1}
,
then
f(x)
is
acyclotom
ic
polynom
ial.
TypicalD
iscussion:
Mahler
measure
Lehm
er’s
polynom
ial
Etc.
Suppose
f(x)
2Z[x]and
|f(0)|
=1.
Know
:f(x)
has
aroot
in
D=
{z
2C
:|z|
1}.
Pictures
ofother
regions
thatm
ustcontain
arootoff(x).
Where
did
these
regions
com
efrom
?
Why
are
they
ofinterest
to
me?
What
are
or
are
not
other
possible
regions
ofinterest?
Supposef(x)2Z[x]and|f(0)|=1.
Know:f(x)hasarootinD={z2C:|z|1}.
Picturesofotherregionsthatmustcontainarootoff(x).
Wheredidtheseregionscomefrom?
Whyaretheyofinteresttome?
Whatareorarenototherpossibleregionsofinterest?
arootinhere
=)Suppose
f(x)
2Z[x]and
|f(0)|
=1.
Know
:f(x)
has
aroot
in
D=
{z
2C
:|z|
1}.
Pictures
ofother
regions
thatm
ustcontain
arootoff(x).
Where
did
these
regions
com
efrom
?
Why
are
they
ofinterest
to
me?
What
are
or
are
not
other
possible
regions
ofinterest?
aroot
out
here
f(x)
monic
=)
Suppose
f(x)
2Z[x]and
|f(0)|
=1.
Know
:f(x)
has
aroot
in
D=
{z
2C
:|z|
1}.
Pictures
ofother
regions
thatm
ustcontain
arootoff(x).
Where
did
these
regions
com
efrom
?
Why
are
they
ofinterest
to
me?
What
are
or
are
not
other
possible
regions
ofinterest?
aroot
in
here
f(x)
monic
=)
Suppose
f(x)
2Z[x]and
|f(0)|
=1.
Know
:f(x)
has
aroot
in
D=
{z
2C
:|z|
1}.
Pictures
ofother
regions
thatm
ustcontain
arootoff(x).
Where
did
these
regions
com
efrom
?
Why
are
they
ofinterest
to
me?
What
are
or
are
not
other
possible
regions
ofinterest?
Suppose
f(x)
2Z[x]is
not
constant
and
|f(0)|
=1.
Know
:f(x)
has
aroot
in
D=
{z
2C
:|z|
1}.
Pictures
ofother
regions
thatm
ustcontain
arootoff(x).
Where
did
these
regions
com
efrom
?
Why
are
they
ofinterest
to
me?
What
are
or
are
not
other
possible
regions
ofinterest?
aroot
out
here
f(x)
monic
=)
Sim
ple
ClassicalR
esult.
Let
f(x)
be
monic
in
C[x]with
deg
f�
1and
|f(0)|
�1.
Then
f(x)
has
aroot
on
or
outside
the
unitcircle
C=
{z
2C
:|z|
=1}
.
Corollary.
Every
non-constant
monic
f(x)
2Z[x]
with
f(0)
6=0
has
arooton
or
outside
C=
{z
2C
:|z|
=1}
.
Theorem
(K
ronecker).
If
f(x)
2Z[x]
is
monic,
irre-
ducible,
and
has
allits
roots
on
C=
{z
2C
:|z|
=1}
,
then
f(x)
is
acyclotom
ic
polynom
ial.
TypicalD
iscussion:
Mahler
measure
Lehm
er’s
polynom
ial
Etc.
Suppose
f(x)
2Z[x]and
|f(0)|
=1.
Know
:f(x)
has
aroot
in
D=
{z
2C
:|z|
1}.
Pictures
ofother
regions
thatm
ustcontain
arootoff(x).
Where
did
these
regions
com
efrom
?
Why
are
they
ofinterest
to
me?
What
are
or
are
not
other
possible
regions
ofinterest?
Supposef(x)2Z[x]and|f(0)|=1.
Know:f(x)hasarootinD={z2C:|z|1}.
Picturesofotherregionsthatmustcontainarootoff(x).
Wheredidtheseregionscomefrom?
Whyaretheyofinteresttome?
Whatareorarenototherpossibleregionsofinterest?
arootinhere
=)
Suppose
f(x)
2Z[x]and
|f(0)|
=1.
Know
:f(x)
has
aroot
in
D=
{z
2C
:|z|
1}.
Pictures
ofother
regions
thatm
ustcontain
arootoff(x).
Where
did
these
regions
com
efrom
?
Why
are
they
ofinterest
to
me?
What
are
or
are
not
other
possible
regions
ofinterest?
aroot
out
here
f(x)
monic
=)
Suppose
f(x)
2Z[x]and
|f(0)|
=1.
Know
:f(x)
has
aroot
in
D=
{z
2C
:|z|
1}.
Pictures
ofother
regions
thatm
ustcontain
arootoff(x).
Where
did
these
regions
com
efrom
?
Why
are
they
ofinterest
to
me?
What
are
or
are
not
other
possible
regions
ofinterest?
aroot
in
here
f(x)
monic
=)
Suppose
f(x)
2Z[x]and
|f(0)|
=1.
Know
:f(x)
has
aroot
in
D=
{z
2C
:|z|
1}.
Pictures
ofother
regions
thatm
ustcontain
arootoff(x).
Where
did
these
regions
com
efrom
?
Why
are
they
ofinterest
to
me?
What
are
or
are
not
other
possible
regions
ofinterest?
Suppose
f(x)
2Z[x]is
not
constant
and
|f(0)|
=1.
Know
:f(x)
has
aroot
in
D=
{z
2C
:|z|
1}.
Pictures
ofother
regions
thatm
ustcontain
arootoff(x).
Where
did
these
regions
com
efrom
?
Why
are
they
ofinterest
to
me?
What
are
or
are
not
other
possible
regions
ofinterest?
aroot
out
here
f(x)
monic
=)
Sim
ple
ClassicalR
esult.
Let
f(x)
be
monic
in
C[x]with
deg
f�
1and
|f(0)|
�1.
Then
f(x)
has
aroot
on
or
outside
the
unitcircle
C=
{z
2C
:|z|
=1}
.
Corollary.
Every
non-constant
monic
f(x)
2Z[x]
with
f(0)
6=0
has
arooton
or
outside
C=
{z
2C
:|z|
=1}
.
Theorem
(K
ronecker).
If
f(x)
2Z[x]
is
monic,
irre-
ducible,
and
has
allits
roots
on
C=
{z
2C
:|z|
=1}
,
then
f(x)
is
acyclotom
ic
polynom
ial.
TypicalD
iscussion:
Mahler
measure
Lehm
er’s
polynom
ial
Etc.
Can
we
say
som
ething
di↵erent
about
f(x)
2Z[x]?
Question:
If
f(x)
2Z[x]is
monic,
irreducible
and
non-cyclotom
ic,
then
there
must
be
aroot
of
f(x)
having
what
distance
from
the
boundary
ofR
?
RCom
ment.Every
non-constantm
onic
polynom
ialf(x)
2Z[x]has
arooton
or
outside
the
region
R.
Can
we
say
som
ething
di↵erent
about
f(x)
2Z[x]?
Question:
If
f(x)
2Z[x]is
monic,
irreducible
and
non-cyclotom
ic,
then
there
must
be
aroot
of
f(x)
having
what
distance
from
the
boundary
ofR
?
Question:
Is
there
som
ething
like
aM
ahler
measure
for
R?
Question:
Is
there
aLehm
er
type
polynom
ialassociated
to
it?
RCom
ment.Every
non-constantm
onic
polynom
ialf(x)
2Z[x]has
arooton
or
outside
the
region
R.
Can
we
say
som
ething
di↵erent
about
f(x)
2Z[x]?
Question:
If
f(x)
2Z[x]is
monic,
irreducible
and
non-cyclotom
ic,
then
there
must
be
aroot
of
f(x)
having
what
distance
from
the
boundary
ofR
?
Question:
Is
there
som
ething
like
aM
ahler
measure
for
R?
Question:
Is
there
aLehm
er
type
polynom
ialassociated
to
it?
RCom
ment.Every
non-constantm
onic
polynom
ialf(x)
2Z[x]has
arooton
or
outside
the
region
R.
Can
we
say
som
ething
di↵erent
about
f(x)
2Z[x]?
Question:
If
f(x)
2Z[x]
is
monic,
irreducible
and
non-
cyclotom
ic,then
there
mustbe
arootoff(x)having
what
distance
from
the
boundary
ofR
?
RCom
ment.Every
non-constantm
onic
polynom
ialf(x)
2Z[x]has
arooton
or
outside
the
region
R.
Why
are
they
ofinterest
to
me?
What
is
the
maxim
um
area
for
such
aregion,
sym
metric
about
the
realaxis?
Sim
ple
ClassicalR
esult.Let
f(x)
=
Pnj=
0
a
j
x
j2C[x]with
n�
1and
|a
n |�|a
0 |>
0.
Then
f(x)
has
aroot
in
the
unitdisk
D=
{z
2C
:|z|
1}.
Proof.
f(x)
=a
n
n
Yj=
1
(x
�↵
j
)=)
|a
0 |=
|a
n ||↵
1 |···|↵
n |
|a
n |�|a
0 |>
0=)
9j
such
that
|↵
j |1
Corollary.
Every
non-constant
polynom
ial
f(x)
2C[x]
with
leading
coe�
cient
�1
and
with
|f(0)|
=1
has
a
rootin
D=
{z
2C
:|z|
1}.
Corollary.
Every
non-constant
polynom
ial
f(x)
2Z[x]
with
|f(0)|
=1
has
arootin
D=
{z
2C
:|z|
1}.
Question.
Can
we
say
more
if
f(x)
2Z[x]?
Yes.
Yes
YES
Sim
ple
ClassicalR
esult.
Let
f(x)
be
monic
in
C[x]with
deg
f�
1and
|f(0)|
�1.
Then
f(x)
has
aroot
on
or
outside
the
unitcircle
C=
{z
2C
:|z|
=1}
.
Corollary.
Every
non-constantm
onic
polynom
ial
f(x)
2Z[x]has
arooton
or
outside
D=
{z
2C
:|z|
=1}
.
Theorem
(K
ronecker).
If
f(x)
2Z[x]
is
monic,
irre-
ducible,and
has
allits
roots
on
D=
{z
2C
:|z|
=1}
,
then
f(x)
is
acyclotom
ic
polynom
ial.
TypicalD
iscussion:
Mahler
measure
Lehm
er’s
polynom
ial
Etc.
Sim
ple
ClassicalR
esult.Let
f(x)
=
Pnj=
0
a
j
x
j2C[x]with
n�
1and
|a
n |�|a
0 |>
0.
Then
f(x)
has
aroot
in
the
unitdisk
D=
{z
2C
:|z|
1}.
Proof.
f(x)
=a
n
n
Yj=
1
(x
�↵
j
)=)
|a
0 |=
|a
n ||↵
1 |···|↵
n |
|a
n |�|a
0 |>
0=)
9j
such
that
|↵
j |1
Corollary.
Every
non-constant
polynom
ial
f(x)
2C[x]
with
leading
coe�
cient
�1
and
with
|f(0)|
=1
has
a
rootin
D=
{z
2C
:|z|
1}.
Corollary.
Every
non-constant
polynom
ial
f(x)
2Z[x]
with
|f(0)|
=1
has
arootin
D=
{z
2C
:|z|
1}.
Question.
Can
we
say
more
if
f(x)
2Z[x]?
Yes.
Yes
YES
Sim
ple
ClassicalR
esult.Let
f(x)
=
Pnj=
0
a
j
x
j2C[x]with
n�
1and
|a
n |�|a
0 |>
0.
Then
f(x)
has
aroot
in
the
unitdisk
D=
{z
2C
:|z|
1}.
Proof.
f(x)
=a
n
n
Yj=
1
(x
�↵
j
)=)
|a
0 |=
|a
n ||↵
1 |···|↵
n |
|a
n |�|a
0 |>
0=)
9j
such
that
|↵
j |1
Corollary.
Every
non-constant
polynom
ial
f(x)
2C[x]
with
leading
coe�
cient
�1
and
with
|f(0)|
=1
has
a
rootin
D=
{z
2C
:|z|
1}.
Corollary.
Every
non-constant
polynom
ial
f(x)
2Z[x]
with
|f(0)|
=1
has
arootin
D=
{z
2C
:|z|
1}.
Question.
Can
we
say
more
if
f(x)
2Z[x]?
Yes.
Yes
YES
Sim
ple
ClassicalR
esult.
Let
f(x)
be
monic
in
C[x]with
deg
f�
1and
|f(0)|
�1.
Then
f(x)
has
aroot
on
or
outside
the
unitcircle
C=
{z
2C
:|z|
=1}
.
Corollary.
Every
non-constant
monic
f(x)
2Z[x]
with
f(0)
6=0
has
arooton
or
outside
C=
{z
2C
:|z|
=1}
.
Theorem
(K
ronecker).
If
f(x)
2Z[x]
is
monic,
irre-
ducible,
and
has
allits
roots
on
C=
{z
2C
:|z|
=1}
,
then
f(x)
is
acyclotom
ic
polynom
ial.
TypicalD
iscussion:
Mahler
measure
Lehm
er’s
polynom
ial
Etc.
Suppose
f(x)
2Z[x]is
not
constant
and
|f(0)|
=1.
Know
:f(x)
has
aroot
in
D=
{z
2C
:|z|
1}.
Pictures
ofother
regions
thatm
ustcontain
arootoff(x).
Where
did
these
regions
com
efrom
?
Why
are
they
ofinterest
to
me?
What
are
or
are
not
other
possible
regions
ofinterest?
aroot
out
here
f(x)
monic
=)
Suppose
f(x)
2Z[x]and
|f(0)|
=1.
Know
:f(x)
has
aroot
in
D=
{z
2C
:|z|
1}.
Pictures
ofother
regions
thatm
ustcontain
arootoff(x).
Where
did
these
regions
com
efrom
?
Why
are
they
ofinterest
to
me?
What
are
or
are
not
other
possible
regions
ofinterest?
Suppose
f(x)
2Z[x]and
|f(0)|
=1.
Know
:f(x)
has
aroot
in
D=
{z
2C
:|z|
1}.
Pictures
ofother
regions
thatm
ustcontain
arootoff(x).
Where
did
these
regions
com
efrom
?
Why
are
they
ofinterest
to
me?
What
are
or
are
not
other
possible
regions
ofinterest?
Suppose
f(x)
2Z[x]and
|f(0)|
=1.
Know
:f(x)
has
aroot
in
D=
{z
2C
:|z|
1}.
Pictures
ofother
regions
thatm
ustcontain
arootoff(x).
Where
did
these
regions
com
efrom
?
Why
are
they
ofinterest
to
me?
What
are
or
are
not
other
possible
regions
ofinterest?
Suppose
f(x)
2Z[x]and
|f(0)|
=1.
Know
:f(x)
has
aroot
in
D=
{z
2C
:|z|
1}.
Pictures
ofother
regions
thatm
ustcontain
arootoff(x).
Where
did
these
regions
com
efrom
?
Why
are
they
ofinterest
to
me?
What
are
or
are
not
other
possible
regions
ofinterest?
Suppose
f(x)
2Z[x]is
not
constant
and
|f(0)|
=1.
Know
:f(x)
has
aroot
in
D=
{z
2C
:|z|
1}.
Pictures
ofother
regions
thatm
ustcontain
arootoff(x).
Where
did
these
regions
com
efrom
?
Why
are
they
ofinterest
to
me?
What
are
or
are
not
other
possible
regions
ofinterest?
aroot
out
here
f(x)
monic
=)
Suppose
f(x)
2Z[x]and
|f(0)|
=1.
Know
:f(x)
has
aroot
in
D=
{z
2C
:|z|
1}.
Pictures
ofother
regions
thatm
ustcontain
arootoff(x).
Where
did
these
regions
com
efrom
?
Why
are
they
ofinterest
to
me?
What
are
or
are
not
other
possible
regions
ofinterest?
Suppose
f(x)
2Z[x]and
|f(0)|
=1.
Know
:f(x)
has
aroot
in
D=
{z
2C
:|z|
1}.
Pictures
ofother
regions
thatm
ustcontain
arootoff(x).
Where
did
these
regions
com
efrom
?
Why
are
they
ofinterest
to
me?
What
are
or
are
not
other
possible
regions
ofinterest?
Suppose
f(x)
2Z[x]and
|f(0)|
=1.
Know
:f(x)
has
aroot
in
D=
{z
2C
:|z|
1}.
Pictures
ofother
regions
thatm
ustcontain
arootoff(x).
Where
did
these
regions
com
efrom
?
Why
are
they
ofinterest
to
me?
What
are
or
are
not
other
possible
regions
ofinterest?
Suppose
f(x)
2Z[x]and
|f(0)|
=1.
Know
:f(x)
has
aroot
in
D=
{z
2C
:|z|
1}.
Pictures
ofother
regions
thatm
ustcontain
arootoff(x).
Where
did
these
regions
com
efrom
?
Why
are
they
ofinterest
to
me?
What
are
or
are
not
other
possible
regions
ofinterest?
aroot
in
here
Supposef(x)2Z[x]and|f(0)|=1.
Know:f(x)hasarootinD={z2C:|z|1}.
Picturesofotherregionsthatmustcontainarootoff(x).
Wheredidtheseregionscomefrom?
Whyaretheyofinteresttome?
Whatareorarenototherpossibleregionsofinterest?
arootinhere
=)
Suppose
f(x)
2Z[x]is
not
constant
and
|f(0)|
=1.
Know
:f(x)
has
aroot
in
D=
{z
2C
:|z|
1}.
Pictures
ofother
regions
thatm
ustcontain
arootoff(x).
Where
did
these
regions
com
efrom
?
Why
are
they
ofinterest
to
me?
What
are
or
are
not
other
possible
regions
ofinterest?
aroot
out
here
f(x)
monic
=)
Suppose
f(x)
2Z[x]and
|f(0)|
=1.
Know
:f(x)
has
aroot
in
D=
{z
2C
:|z|
1}.
Pictures
ofother
regions
thatm
ustcontain
arootoff(x).
Where
did
these
regions
com
efrom
?
Why
are
they
ofinterest
to
me?
What
are
or
are
not
other
possible
regions
ofinterest?
Suppose
f(x)
2Z[x]and
|f(0)|
=1.
Know
:f(x)
has
aroot
in
D=
{z
2C
:|z|
1}.
Pictures
ofother
regions
thatm
ustcontain
arootoff(x).
Where
did
these
regions
com
efrom
?
Why
are
they
ofinterest
to
me?
What
are
or
are
not
other
possible
regions
ofinterest?
aroot
in
here
Supposef(x)2Z[x]and|f(0)|=1.
Know:f(x)hasarootinD={z2C:|z|1}.
Picturesofotherregionsthatmustcontainarootoff(x).
Wheredidtheseregionscomefrom?
Whyaretheyofinteresttome?
Whatareorarenototherpossibleregionsofinterest?
arootinhere
=)
Suppose
f(x)
2Z[x]and
|f(0)|
=1.
Know
:f(x)
has
aroot
in
D=
{z
2C
:|z|
1}.
Pictures
ofother
regions
thatm
ustcontain
arootoff(x).
Where
did
these
regions
com
efrom
?
Why
are
they
ofinterest
to
me?
What
are
or
are
not
other
possible
regions
ofinterest?
Suppose
f(x)
2Z[x]and
|f(0)|
=1.
Know
:f(x)
has
aroot
in
D=
{z
2C
:|z|
1}.
Pictures
ofother
regions
thatm
ustcontain
arootoff(x).
Where
did
these
regions
com
efrom
?
Why
are
they
ofinterest
to
me?
What
are
or
are
not
other
possible
regions
ofinterest?
Suppose
f(x)
2Z[x]is
not
constant
and
|f(0)|
=1.
Know
:f(x)
has
aroot
in
D=
{z
2C
:|z|
1}.
Pictures
ofother
regions
thatm
ustcontain
arootoff(x).
Where
did
these
regions
com
efrom
?
Why
are
they
ofinterest
to
me?
What
are
or
are
not
other
possible
regions
ofinterest?
aroot
out
here
f(x)
monic
=)
Suppose
f(x)
2Z[x]and
|f(0)|
=1.
Know
:f(x)
has
aroot
in
D=
{z
2C
:|z|
1}.
Pictures
ofother
regions
thatm
ustcontain
arootoff(x).
Where
did
these
regions
com
efrom
?
Why
are
they
ofinterest
to
me?
What
are
or
are
not
other
possible
regions
ofinterest?
Suppose
f(x)
2Z[x]and
|f(0)|
=1.
Know
:f(x)
has
aroot
in
D=
{z
2C
:|z|
1}.
Pictures
ofother
regions
thatm
ustcontain
arootoff(x).
Where
did
these
regions
com
efrom
?
Why
are
they
ofinterest
to
me?
What
are
or
are
not
other
possible
regions
ofinterest?
aroot
in
here
Supposef(x)2Z[x]and|f(0)|=1.
Know:f(x)hasarootinD={z2C:|z|1}.
Picturesofotherregionsthatmustcontainarootoff(x).
Wheredidtheseregionscomefrom?
Whyaretheyofinteresttome?
Whatareorarenototherpossibleregionsofinterest?
arootinhere
=)
Suppose
f(x)
2Z[x]and
|f(0)|
=1.
Know
:f(x)
has
aroot
in
D=
{z
2C
:|z|
1}.
Pictures
ofother
regions
thatm
ustcontain
arootoff(x).
Where
did
these
regions
com
efrom
?
Why
are
they
ofinterest
to
me?
What
are
or
are
not
other
possible
regions
ofinterest?
Suppose
f(x)
2Z[x]and
|f(0)|
=1.
Know
:f(x)
has
aroot
in
D=
{z
2C
:|z|
1}.
Pictures
ofother
regions
thatm
ustcontain
arootoff(x).
Where
did
these
regions
com
efrom
?
Why
are
they
ofinterest
to
me?
What
are
or
are
not
other
possible
regions
ofinterest?
Suppose
f(x)
2Z[x]is
not
constant
and
|f(0)|
=1.
Know
:f(x)
has
aroot
in
D=
{z
2C
:|z|
1}.
Pictures
ofother
regions
thatm
ustcontain
arootoff(x).
Where
did
these
regions
com
efrom
?
Why
are
they
ofinterest
to
me?
What
are
or
are
not
other
possible
regions
ofinterest?
aroot
out
here
f(x)
monic
=)
Suppose
f(x)
2Z[x]and
|f(0)|
=1.
Know
:f(x)
has
aroot
in
D=
{z
2C
:|z|
1}.
Pictures
ofother
regions
thatm
ustcontain
arootoff(x).
Where
did
these
regions
com
efrom
?
Why
are
they
ofinterest
to
me?
What
are
or
are
not
other
possible
regions
ofinterest?
Suppose
f(x)
2Z[x]and
|f(0)|
=1.
Know
:f(x)
has
aroot
in
D=
{z
2C
:|z|
1}.
Pictures
ofother
regions
thatm
ustcontain
arootoff(x).
Where
did
these
regions
com
efrom
?
Why
are
they
ofinterest
to
me?
What
are
or
are
not
other
possible
regions
ofinterest?
aroot
in
here
Supposef(x)2Z[x]and|f(0)|=1.
Know:f(x)hasarootinD={z2C:|z|1}.
Picturesofotherregionsthatmustcontainarootoff(x).
Wheredidtheseregionscomefrom?
Whyaretheyofinteresttome?
Whatareorarenototherpossibleregionsofinterest?
arootinhere
=)Suppose
f(x)
2Z[x]and
|f(0)|
=1.
Know
:f(x)
has
aroot
in
D=
{z
2C
:|z|
1}.
Pictures
ofother
regions
thatm
ustcontain
arootoff(x).
Where
did
these
regions
com
efrom
?
Why
are
they
ofinterest
to
me?
What
are
or
are
not
other
possible
regions
ofinterest?
Suppose
f(x)
2Z[x]and
|f(0)|
=1.
Know
:f(x)
has
aroot
in
D=
{z
2C
:|z|
1}.
Pictures
ofother
regions
thatm
ustcontain
arootoff(x).
Where
did
these
regions
com
efrom
?
Why
are
they
ofinterest
to
me?
What
are
or
are
not
other
possible
regions
ofinterest?
Suppose
f(x)
2Z[x]is
not
constant
and
|f(0)|
=1.
Know
:f(x)
has
aroot
in
D=
{z
2C
:|z|
1}.
Pictures
ofother
regions
thatm
ustcontain
arootoff(x).
Where
did
these
regions
com
efrom
?
Why
are
they
ofinterest
to
me?
What
are
or
are
not
other
possible
regions
ofinterest?
aroot
out
here
f(x)
monic
=)
Suppose
f(x)
2Z[x]and
|f(0)|
=1.
Know
:f(x)
has
aroot
in
D=
{z
2C
:|z|
1}.
Pictures
ofother
regions
thatm
ustcontain
arootoff(x).
Where
did
these
regions
com
efrom
?
Why
are
they
ofinterest
to
me?
What
are
or
are
not
other
possible
regions
ofinterest?
Suppose
f(x)
2Z[x]and
|f(0)|
=1.
Know
:f(x)
has
aroot
in
D=
{z
2C
:|z|
1}.
Pictures
ofother
regions
thatm
ustcontain
arootoff(x).
Where
did
these
regions
com
efrom
?
Why
are
they
ofinterest
to
me?
What
are
or
are
not
other
possible
regions
ofinterest?
aroot
in
here
Supposef(x)2Z[x]and|f(0)|=1.
Know:f(x)hasarootinD={z2C:|z|1}.
Picturesofotherregionsthatmustcontainarootoff(x).
Wheredidtheseregionscomefrom?
Whyaretheyofinteresttome?
Whatareorarenototherpossibleregionsofinterest?
arootinhere
=)
Suppose
f(x)
2Z[x]and
|f(0)|
=1.
Know
:f(x)
has
aroot
in
D=
{z
2C
:|z|
1}.
Pictures
ofother
regions
thatm
ustcontain
arootoff(x).
Where
did
these
regions
com
efrom
?
Why
are
they
ofinterest
to
me?
What
are
or
are
not
other
possible
regions
ofinterest?
Suppose
f(x)
2Z[x]and
|f(0)|
=1.
Know
:f(x)
has
aroot
in
D=
{z
2C
:|z|
1}.
Pictures
ofother
regions
thatm
ustcontain
arootoff(x).
Where
did
these
regions
com
efrom
?
Why
are
they
ofinterest
to
me?
What
are
or
are
not
other
possible
regions
ofinterest?
Suppose
f(x)
2Z[x]is
not
constant
and
|f(0)|
=1.
Know
:f(x)
has
aroot
in
D=
{z
2C
:|z|
1}.
Pictures
ofother
regions
thatm
ustcontain
arootoff(x).
Where
did
these
regions
com
efrom
?
Why
are
they
ofinterest
to
me?
What
are
or
are
not
other
possible
regions
ofinterest?
aroot
out
here
f(x)
monic
=)
Suppose
f(x)
2Z[x]and
|f(0)|
=1.
Know
:f(x)
has
aroot
in
D=
{z
2C
:|z|
1}.
Pictures
ofother
regions
thatm
ustcontain
arootoff(x).
Where
did
these
regions
com
efrom
?
Why
are
they
ofinterest
to
me?
What
are
or
are
not
other
possible
regions
ofinterest?
Suppose
f(x)
2Z[x]and
|f(0)|
=1.
Know
:f(x)
has
aroot
in
D=
{z
2C
:|z|
1}.
Pictures
ofother
regions
thatm
ustcontain
arootoff(x).
Where
did
these
regions
com
efrom
?
Why
are
they
ofinterest
to
me?
What
are
or
are
not
other
possible
regions
ofinterest?
aroot
in
here
Supposef(x)2Z[x]and|f(0)|=1.
Know:f(x)hasarootinD={z2C:|z|1}.
Picturesofotherregionsthatmustcontainarootoff(x).
Wheredidtheseregionscomefrom?
Whyaretheyofinteresttome?
Whatareorarenototherpossibleregionsofinterest?
arootinhere
=)
Suppose
f(x)
2Z[x]and
|f(0)|
=1.
Know
:f(x)
has
aroot
in
D=
{z
2C
:|z|
1}.
Pictures
ofother
regions
thatm
ustcontain
arootoff(x).
Where
did
these
regions
com
efrom
?
Why
are
they
ofinterest
to
me?
What
are
or
are
not
other
possible
regions
ofinterest?
Suppose
f(x)
2Z[x]and
|f(0)|
=1.
Know
:f(x)
has
aroot
in
D=
{z
2C
:|z|
1}.
Pictures
ofother
regions
thatm
ustcontain
arootoff(x).
Where
did
these
regions
com
efrom
?
Why
are
they
ofinterest
to
me?
What
are
or
are
not
other
possible
regions
ofinterest?
Suppose
f(x)
2Z[x]is
not
constant
and
|f(0)|
=1.
Know
:f(x)
has
aroot
in
D=
{z
2C
:|z|
1}.
Pictures
ofother
regions
thatm
ustcontain
arootoff(x).
Where
did
these
regions
com
efrom
?
Why
are
they
ofinterest
to
me?
What
are
or
are
not
other
possible
regions
ofinterest?
aroot
out
here
f(x)
monic
=)
Suppose
f(x)
2Z[x]and
|f(0)|
=1.
Know
:f(x)
has
aroot
in
D=
{z
2C
:|z|
1}.
Pictures
ofother
regions
thatm
ustcontain
arootoff(x).
Where
did
these
regions
com
efrom
?
Why
are
they
ofinterest
to
me?
What
are
or
are
not
other
possible
regions
ofinterest?
aroot
in
here
f(x)
monic
=)
Suppose
f(x)
2Z[x]and
|f(0)|
=1.
Know
:f(x)
has
aroot
in
D=
{z
2C
:|z|
1}.
Pictures
ofother
regions
thatm
ustcontain
arootoff(x).
Where
did
these
regions
com
efrom
?
Why
are
they
ofinterest
to
me?
What
are
or
are
not
other
possible
regions
ofinterest?
Supposef(x)2Z[x]and|f(0)|=1.
Know:f(x)hasarootinD={z2C:|z|1}.
Picturesofotherregionsthatmustcontainarootoff(x).
Wheredidtheseregionscomefrom?
Whyaretheyofinteresttome?
Whatareorarenototherpossibleregionsofinterest?
arootinhere
=)Suppose
f(x)
2Z[x]and
|f(0)|
=1.
Know
:f(x)
has
aroot
in
D=
{z
2C
:|z|
1}.
Pictures
ofother
regions
thatm
ustcontain
arootoff(x).
Where
did
these
regions
com
efrom
?
Why
are
they
ofinterest
to
me?
What
are
or
are
not
other
possible
regions
ofinterest?
aroot
out
here
f(x)
monic
=)
Suppose
f(x)
2Z[x]and
|f(0)|
=1.
Know
:f(x)
has
aroot
in
D=
{z
2C
:|z|
1}.
Pictures
ofother
regions
thatm
ustcontain
arootoff(x).
Where
did
these
regions
com
efrom
?
Why
are
they
ofinterest
to
me?
What
are
or
are
not
other
possible
regions
ofinterest?
Suppose
f(x)
2Z[x]is
not
constant
and
|f(0)|
=1.
Know
:f(x)
has
aroot
in
D=
{z
2C
:|z|
1}.
Pictures
ofother
regions
thatm
ustcontain
arootoff(x).
Where
did
these
regions
com
efrom
?
Why
are
they
ofinterest
to
me?
What
are
or
are
not
other
possible
regions
ofinterest?
aroot
out
here
f(x)
monic
=)
Sim
ple
ClassicalR
esult.
Let
f(x)
be
monic
in
C[x]with
deg
f�
1and
|f(0)|
�1.
Then
f(x)
has
aroot
on
or
outside
the
unitcircle
C=
{z
2C
:|z|
=1}
.
Corollary.
Every
non-constant
monic
f(x)
2Z[x]
with
f(0)
6=0
has
arooton
or
outside
C=
{z
2C
:|z|
=1}
.
Theorem
(K
ronecker).
If
f(x)
2Z[x]
is
monic,
irre-
ducible,
and
has
allits
roots
on
C=
{z
2C
:|z|
=1}
,
then
f(x)
is
acyclotom
ic
polynom
ial.
TypicalD
iscussion:
Mahler
measure
Lehm
er’s
polynom
ial
Etc.
Where
do
these
regions
com
efrom
?
N
b
(z)
=|b
�1
�z|
2e
1 �|b
+⇣
3 �z||
b+
⇣
3 �z| �
2e
3
·(|
b+
i�z||
b�
i�z|
)
2e
4 �|b
+⇣
6 �z||
b+
⇣
6 �z| �
2e
6,
N(x,y)
=|z
+1|
2e
1 �|z
�⇣
3 ||z
�⇣
3 | �2e
3
·(|
z�
i||z
+i|)
2e
4 �|z
�⇣
6 ||z
�⇣
6 | �2e
6
D(x,y)
=|z|
2d
Let
F(z)
=
N(z)
D(z)
where
N(z)
=|z
+1|
2e
2 �|z
�⇣
3 ||z
�⇣
3 | �2e
3
·(|
z�
i||z
+i|)
2e
4 �|z
�⇣
6 ||z
�⇣
6 | �2e
6
and
D(z)
=|z|
2d
,w
here
d�
e
2
+2e
3
+2e
4
+2e
6
.
f(x)
=a
r
Yj=
1
(x
�↵
j
)1
|a|
m
num
=|f
(�1)|
2e
2|f
(⇣
3
)|2e
3 ��f
�⇣
3 � ��2e
3
·|f
(i)|
2e
4|f
(�i)|
2e
4|f
(⇣
6
)|2e
6 ��f
�⇣
6 � ��2e
6
den
=|f
(0)|
2d
num
den
=
1
|a|
m
r
Yj=
1
F(↵
j
)
z=
x+
iy
F=
N
D
N=
|z
+1|
2e
1 �|z
�⇣
3 ||z
�⇣
3 | �2e
3
·(|
z�
i||z
+i|)
2e
4 �|z
�⇣
6 ||z
�⇣
6 | �2e
6
D=
|z|
2d
,d
�2e
1
+4e
3
+4e
4
+4e
6
.
z=
x+
iy
F=
N
D
N=
|z
+1|
2e
1 �|z
�⇣
3 ||z
�⇣
3 | �2e
3
·(|
z�
i||z
+i|)
2e
4 �|z
�⇣
6 ||z
�⇣
6 | �2e
6
D=
|z|
2d
d�
2e
1
+4e
3
+4e
4
+4e
6
.
f(x)
=a
r
Yj=
1
(x
�↵
j
)1
|a|
m
num
=|f
(�1)|
2e
2|f
(⇣
3
)|2e
3 ��f
�⇣
3 � ��2e
3
·|f
(i)|
2e
4|f
(�i)|
2e
4|f
(⇣
6
)|2e
6 ��f
�⇣
6 � ��2e
6
den
=|f
(0)|
2d
num
den
=
1
|a|
m
r
Yj=
1
F(↵
j
)
f(x)
=a
r
Yj=
1
(x
�↵
j
)1
|a|
m
num
=|f
(�1)|
2e
2|f
(⇣
3
)|2e
3 ��f
�⇣
3 � ��2e
3
·|f
(i)|
2e
4|f
(�i)|
2e
4|f
(⇣
6
)|2e
6 ��f
�⇣
6 � ��2e
6
den
=|f
(0)|
2d
num
den
=
1
|a|
m
r
Yj=
1
F(↵
j
)
z=
x+
iy
F=
N
D
N=
|z
+1|
2e
2 �|z
�⇣
3 ||z
�⇣
3 | �2e
3
·(|
z�
i||z
+i|)
2e
4 �|z
�⇣
6 ||z
�⇣
6 | �2e
6
D=
|z|
2d
d�
2e
2
+4e
3
+4e
4
+4e
6
z=
x+
iy
F=
N
D
N=
|z
+1|
2e
2 �|z
�⇣
3 ||z
�⇣
3 | �2e
3
·(|
z�
i||z
+i|)
2e
4 �|z
�⇣
6 ||z
�⇣
6 | �2e
6
D=
|z|
2d
d�
e
2
+2e
3
+2e
4
+2e
6
f(x)
=a
r
Yj=
1
(x
�↵
j
)1
|a|
m
num
=|f
(�1)|
2e
2|f
(⇣
3
)|2e
3 ��f
�⇣
3 � ��2e
3
·|f
(i)|
2e
4|f
(�i)|
2e
4|f
(⇣
6
)|2e
6 ��f
�⇣
6 � ��2e
6
den
=|f
(0)|
2d
num
den
=
1
|a|
m
r
Yj=
1
F(↵
j
)
z=
x+
iy
F=
N
D
N=
|z
+1|
2e
1 �|z
�⇣
3 ||z
�⇣
3 | �2e
3
·(|
z�
i||z
+i|)
2e
4 �|z
�⇣
6 ||z
�⇣
6 | �2e
6
D=
|z|
2d
,d
�2e
1
+4e
3
+4e
4
+4e
6
.
z=
x+
iy
F=
N
D
N=
|z
+1|
2e
1 �|z
�⇣
3 ||z
�⇣
3 | �2e
3
·(|
z�
i||z
+i|)
2e
4 �|z
�⇣
6 ||z
�⇣
6 | �2e
6
D=
|z|
2d
d�
2e
1
+4e
3
+4e
4
+4e
6
.
f(x)
=a
r
Yj=
1
(x
�↵
j
)1
|a|
m
num
=|f
(�1)|
2e
2|f
(⇣
3
)|2e
3 ��f
�⇣
3 � ��2e
3
·|f
(i)|
2e
4|f
(�i)|
2e
4|f
(⇣
6
)|2e
6 ��f
�⇣
6 � ��2e
6
den
=|f
(0)|
2d
num
den
=
1
|a|
m
r
Yj=
1
F(↵
j
)
f(x)
=a
r
Yj=
1
(x
�↵
j
)1
|a|
m
num
=|f
(�1)|
2e
2|f
(⇣
3
)|2e
3 ��f
�⇣
3 � ��2e
3
·|f
(i)|
2e
4|f
(�i)|
2e
4|f
(⇣
6
)|2e
6 ��f
�⇣
6 � ��2e
6
den
=|f
(0)|
2d
num
den
=
1
|a|
m
r
Yj=
1
F(↵
j
)
z=
x+
iy
F=
N
D
N=
|z
+1|
2e
2 �|z
�⇣
3 ||z
�⇣
3 | �2e
3
·(|
z�
i||z
+i|)
2e
4 �|z
�⇣
6 ||z
�⇣
6 | �2e
6
D=
|z|
2d
d�
2e
2
+4e
3
+4e
4
+4e
6
z=
x+
iy
F=
N
D
N=
|z
+1|
2e
2 �|z
�⇣
3 ||z
�⇣
3 | �2e
3
·(|
z�
i||z
+i|)
2e
4 �|z
�⇣
6 ||z
�⇣
6 | �2e
6
D=
|z|
2d
d�
e
2
+2e
3
+2e
4
+2e
6
=|a| 2
dr
Yj=
1 |↵j | 2
d
=|a| 2
e2
rYj=
1 |↵j+
1| 2e2
=|a| 2
e3
rYj=
1 |↵j �
⇣3 | 2
e3
=|a| 2
dr
Yj=
1 |↵j | 2
d
|a| 2e2
rYj=
1 |↵j+
1| 2e2
|a| 2e3
rYj=
1 |↵j �
⇣3 | 2
e3
=|a| 2
dr
Yj=
1 |↵j | 2
d
|a| 2e2
rYj=
1 |↵j+
1| 2e2
|a| 2e3
rYj=
1 |↵j �
⇣3 | 2
e3
|a| 2e3
rYj=
1 |↵j �
⇣3 | 2
e3
f(x)
=a
r
Yj=
1
(x
�↵
j
)1
|a|
m
num
=|f
(�1)|
2e
2|f
(⇣
3
)|2e
3 ��f
�⇣
3 � ��2e
3
·|f
(i)|
2e
4|f
(�i)|
2e
4|f
(⇣
6
)|2e
6 ��f
�⇣
6 � ��2e
6
den
=|f
(0)|
2d
num
den
=
1
|a|
m
r
Yj=
1
F(↵
j
)
f(x)
=a
r
Yj=
1
(x
�↵
j
)1
|a|
m
num
=|f
(�1)|
2e
2|f
(⇣
3
)|2e
3 ��f
�⇣
3 � ��2e
3
·|f
(i)|
2e
4|f
(�i)|
2e
4|f
(⇣
6
)|2e
6 ��f
�⇣
6 � ��2e
6
den
=|f
(0)|
2d
num
den
=
1
|a|
m
r
Yj=
1
F(↵
j
)
f(x)
=a
r
Yj=
1
(x
�↵
j
)1
|a|
m
num
=|f
(�1)|
2e
2|f
(⇣
3
)|2e
3 ��f
�⇣
3 � ��2e
3
·|f
(i)|
2e
4|f
(�i)|
2e
4|f
(⇣
6
)|2e
6 ��f
�⇣
6 � ��2e
6
den
=|f
(0)|
2d
num
den
=
1
|a|
m
r
Yj=
1
F(↵
j
)
f(x)
=a
r
Yj=
1
(x
�↵
j
)1
|a|
m
num
=|f
(�1)|
2e
2|f
(⇣
3
)|2e
3 ��f
�⇣
3 � ��2e
3
·|f
(i)|
2e
4|f
(�i)|
2e
4|f
(⇣
6
)|2e
6 ��f
�⇣
6 � ��2e
6
den
=|f
(0)|
2d
num
den
=
1
|a|
m
r
Yj=
1
F(↵
j
)
f(x)
=a
r
Yj=
1
(x
�↵
j
)1
|a|
m
num
=|f
(�1)|
2e
2|f
(⇣
3
)|2e
3 ��f
�⇣
3 � ��2e
3
·|f
(i)|
2e
4|f
(�i)|
2e
4|f
(⇣
6
)|2e
6 ��f
�⇣
6 � ��2e
6
den
=|f
(0)|
2d
num
den
=
1
|a|
m
r
Yj=
1
F(↵
j
)
f(x)
=a
r
Yj=
1
(x
�↵
j
)1
|a|
m
num
=|f
(�1)|
2e
2|f
(⇣
3
)|2e
3 ��f
�⇣
3 � ��2e
3
·|f
(i)|
2e
4|f
(�i)|
2e
4|f
(⇣
6
)|2e
6 ��f
�⇣
6 � ��2e
6
den
=|f
(0)|
2d
num
den
=
1
|a|
m
r
Yj=
1
F(↵
j
)
f(x)
=a
r
Yj=
1
(x
�↵
j
)1
|a|
m
num
=|f
(�1)|
2e
2|f
(⇣
3
)|2e
3 ��f
�⇣
3 � ��2e
3
·|f
(i)|
2e
4|f
(�i)|
2e
4|f
(⇣
6
)|2e
6 ��f
�⇣
6 � ��2e
6
den
=|f
(0)|
2d
num
den
=
1
|a|
m
r
Yj=
1
F(↵
j
)
f(x)
=a
r
Yj=
1
(x
�↵
j
)1
|a|
m
num
=|f
(�1)|
2e
2|f
(⇣
3
)|2e
3 ��f
�⇣
3 � ��2e
3
·|f
(i)|
2e
4|f
(�i)|
2e
4|f
(⇣
6
)|2e
6 ��f
�⇣
6 � ��2e
6
den
=|f
(0)|
2d
num
den
=
1
|a|
m
r
Yj=
1
F(↵
j
)
Suppose
f(x)
2Z[x]and
|f(0)|
=1.
Know
:f(x)
has
aroot
in
D=
{z
2C
:|z|
1}.
Pictures
ofother
regions
thatm
ustcontain
arootoff(x).
Where
did
these
regions
com
efrom
?
Why
are
they
ofinterest
to
me?
What
are
or
are
not
other
possible
regions
ofinterest?
Suppose
f(x)
2Z[x]and
|f(0)|
=1.
Know
:f(x)
has
aroot
in
D=
{z
2C
:|z|
1}.
Pictures
ofother
regions
thatm
ustcontain
arootoff(x).
Where
did
these
regions
com
efrom
?
Why
are
they
ofinterest
to
me?
What
are
or
are
not
other
possible
regions
ofinterest?
f(x)
=a
r
Yj=
1
(x
�↵
j
)1
|a|
m
num
=|f
(�1)|
2e
2|f
(⇣
3
)|2e
3 ��f
�⇣
3 � ��2e
3
·|f
(i)|
2e
4|f
(�i)|
2e
4|f
(⇣
6
)|2e
6 ��f
�⇣
6 � ��2e
6
den
=|f
(0)|
2d
num
den
=
1
|a|
m
r
Yj=
1
F(↵
j
)
f(x)
=a
r
Yj=
1
(x
�↵
j
)1
|a|
m
num
=|f
(�1)|
2e
2|f
(⇣
3
)|2e
3 ��f
�⇣
3 � ��2e
3
·|f
(i)|
2e
4|f
(�i)|
2e
4|f
(⇣
6
)|2e
6 ��f
�⇣
6 � ��2e
6
den
=|f
(0)|
2d
num
den
=
1
|a|
m
r
Yj=
1
F(↵
j
)
f(x)
=a
r
Yj=
1
(x
�↵
j
)1
|a|
m
num
=|f
(�1)|
2e
2|f
(⇣
3
)|2e
3 ��f
�⇣
3 � ��2e
3
·|f
(i)|
2e
4|f
(�i)|
2e
4|f
(⇣
6
)|2e
6 ��f
�⇣
6 � ��2e
6
den
=|f
(0)|
2d
num
den
=
1
|a|
m
r
Yj=
1
F(↵
j
)
f(x)
=a
r
Yj=
1
(x
�↵
j
)1
|a|
m
num
=|f
(�1)|
2e
2|f
(⇣
3
)|2e
3 ��f
�⇣
3 � ��2e
3
·|f
(i)|
2e
4|f
(�i)|
2e
4|f
(⇣
6
)|2e
6 ��f
�⇣
6 � ��2e
6
den
=|f
(0)|
2d
num
den
=
1
|a|
m
r
Yj=
1
F(↵
j
)
r
Yj=
1
F(↵
j
)�
1
Som
eroot
of
f(x)
is
in
the
set
R=
{z
2C
:F
(z)
�1}
.
R=
{x
+iy
2C
:N
(x,y)
�D
(x,y)}
10
10+
i
19+
p3
i
2
21+
p3
i
2
Suppose
f(x)
2Z[x]and
|f(0)|
=1.
Know
:f(x)
has
aroot
in
D=
{z
2C
:|z|
1}.
Pictures
ofother
regions
thatm
ustcontain
arootoff(x).
Where
did
these
regions
com
efrom
?
Why
are
they
ofinterest
to
me?
What
are
or
are
not
other
possible
regions
ofinterest?
Suppose
f(x)
2Z[x]and
|f(0)|
=1.
Know
:f(x)
has
aroot
in
D=
{z
2C
:|z|
1}.
Pictures
ofother
regions
thatm
ustcontain
arootoff(x).
Where
did
these
regions
com
efrom
?
Why
are
they
ofinterest
to
me?
What
are
or
are
not
other
possible
regions
ofinterest?
Som
eroot
of
f(x)
is
in
the
set
R=
{⇣
2
,⇣
3
,⇣
4
,⇣
6 }[
{z
2C
:F
(z)
�1}
.
R=
{x
+iy
2C
:N
(x,y)
�D
(x,y)}
10
10+
i
19+
p3
i
2
21+
p3
i
2
Som
eroot
of
f(x)
is
in
the
set
R=
{z
2C
:F
(z)
�1}
.
R=
{x
+iy
2C
:N
(x,y)
�D
(x,y)}
10
10+
i
19+
p3
i
2
21+
p3
i
2
Suppose
f(x)
2Z[x]and
|f(0)|
=1.
Know
:f(x)
has
aroot
in
D=
{z
2C
:|z|
1}.
Pictures
ofother
regions
thatm
ustcontain
arootoff(x).
Where
did
these
regions
com
efrom
?
Why
are
they
ofinterest
to
me?
What
are
or
are
not
other
possible
regions
ofinterest?
Suppose
f(x)
2Z[x]and
|f(0)|
=1.
Know
:f(x)
has
aroot
in
D=
{z
2C
:|z|
1}.
Pictures
ofother
regions
thatm
ustcontain
arootoff(x).
Where
did
these
regions
com
efrom
?
Why
are
they
ofinterest
to
me?
What
are
or
are
not
other
possible
regions
ofinterest?
z=
x+
iy
F=
N
D
N=
|z
+1|
2e
1 �|z
�⇣
3 ||z
�⇣
3 | �2e
3
·(|
z�
i||z
+i|)
2e
4 �|z
�⇣
6 ||z
�⇣
6 | �2e
6
D=
|z|
2d
,d
�2e
1
+4e
3
+4e
4
+4e
6
.
z=
x+
iy
F=
N
D
N=
|z
+1|
2e
1 �|z
�⇣
3 ||z
�⇣
3 | �2e
3
·(|
z�
i||z
+i|)
2e
4 �|z
�⇣
6 ||z
�⇣
6 | �2e
6
D=
|z|
2d
d�
2e
1
+4e
3
+4e
4
+4e
6
.
z=
x+
iy
F=
N
D
N=
|z
+1|
2e
2 �|z
�⇣
3 ||z
�⇣
3 | �2e
3
·(|
z�
i||z
+i|)
2e
4 �|z
�⇣
6 ||z
�⇣
6 | �2e
6
D=
|z|
2d
d�
2e
2
+4e
3
+4e
4
+4e
6
z=
x+
iy
F=
N
D
N=
|z
+1|
2e
2 �|z
�⇣
3 ||z
�⇣
3 | �2e
3
·(|
z�
i||z
+i|)
2e
4 �|z
�⇣
6 ||z
�⇣
6 | �2e
6
D=
|z|
2d
d�
e
2
+2e
3
+2e
4
+2e
6
Let
F(z)
=
N(z)
D(z)
where
N(z)
=|z
+1|
2e
2 �|z
�⇣
3 ||z
�⇣
3 | �2e
3
·(|
z�
i||z
+i|)
2e
4 �|z
�⇣
6 ||z
�⇣
6 | �2e
6
and
D(z)
=|z|
2d
,w
here
d�
e
2
+2e
3
+2e
4
+2e
6
.
e
2
=0,
e
3
=4,
e
4
=2,
e
6
=3,
d=
22
Som
eroot
of
f(x)
is
in
the
set
R=
{⇣
2
,⇣
3
,⇣
4
,⇣
6 }[
{z
2C
:F
(z)
�1}
.
R=
{x
+iy
2C
:N
(x,y)
�D
(x,y)}
10
10+
i
19+
p3
i
2
21+
p3
i
2
Som
eroot
of
f(x)
is
in
the
set
R=
{z
2C
:F
(z)
�1}
.
R=
{x
+iy
2C
:N
(x,y)
�D
(x,y)}
10
10+
i
19+
p3
i
2
21+
p3
i
2
Suppose
f(x)
2Z[x]and
|f(0)|
=1.
Know
:f(x)
has
aroot
in
D=
{z
2C
:|z|
1}.
Pictures
ofother
regions
thatm
ustcontain
arootoff(x).
Where
did
these
regions
com
efrom
?
Why
are
they
ofinterest
to
me?
What
are
or
are
not
other
possible
regions
ofinterest?
Suppose
f(x)
2Z[x]and
|f(0)|
=1.
Know
:f(x)
has
aroot
in
D=
{z
2C
:|z|
1}.
Pictures
ofother
regions
thatm
ustcontain
arootoff(x).
Where
did
these
regions
com
efrom
?
Why
are
they
ofinterest
to
me?
What
are
or
are
not
other
possible
regions
ofinterest?
Let
F(z)
=
N(z)
D(z)
where
N(z)
=|z
+1|
2e
2 �|z
�⇣
3 ||z
�⇣
3 | �2e
3
·(|
z�
i||z
+i|)
2e
4 �|z
�⇣
6 ||z
�⇣
6 | �2e
6
and
D(z)
=|z|
2d
,w
here
d�
e
2
+2e
3
+2e
4
+2e
6
.
e
2
=0,
e
3
=4,
e
4
=2,
e
6
=3,
d=
22
Som
eroot
of
f(x)
is
in
the
set
R=
{⇣
2
,⇣
3
,⇣
4
,⇣
6 }[
{z
2C
:F
(z)
�1}
.
R=
{x
+iy
2C
:N
(x,y)
�D
(x,y)}
10
10+
i
19+
p3
i
2
21+
p3
i
2
Suppose
f(x)
2Z[x]and
|f(0)|
=1.
Know
:f(x)
has
aroot
in
D=
{z
2C
:|z|
1}.
Pictures
ofother
regions
thatm
ustcontain
arootoff(x).
Where
did
these
regions
com
efrom
?
Why
are
they
ofinterest
to
me?
What
are
or
are
not
other
possible
regions
ofinterest?
Suppose
f(x)
2Z[x]and
|f(0)|
=1.
Know
:f(x)
has
aroot
in
D=
{z
2C
:|z|
1}.
Pictures
ofother
regions
thatm
ustcontain
arootoff(x).
Where
did
these
regions
com
efrom
?
Why
are
they
ofinterest
to
me?
What
are
or
are
not
other
possible
regions
ofinterest?
aroot
in
here
Supposef(x)2Z[x]and|f(0)|=1.
Know:f(x)hasarootinD={z2C:|z|1}.
Picturesofotherregionsthatmustcontainarootoff(x).
Wheredidtheseregionscomefrom?
Whyaretheyofinteresttome?
Whatareorarenototherpossibleregionsofinterest?
arootinhere
=)Suppose
f(x)
2Z[x]and
|f(0)|
=1.
Know
:f(x)
has
aroot
in
D=
{z
2C
:|z|
1}.
Pictures
ofother
regions
thatm
ustcontain
arootoff(x).
Where
did
these
regions
com
efrom
?
Why
are
they
ofinterest
to
me?
What
are
or
are
not
other
possible
regions
ofinterest?
Suppose
f(x)
2Z[x]and
|f(0)|
=1.
Know
:f(x)
has
aroot
in
D=
{z
2C
:|z|
1}.
Pictures
ofother
regions
thatm
ustcontain
arootoff(x).
Where
did
these
regions
com
efrom
?
Why
are
they
ofinterest
to
me?
What
are
or
are
not
other
possible
regions
ofinterest?
Suppose
f(x)
2Z[x]is
not
constant
and
|f(0)|
=1.
Know
:f(x)
has
aroot
in
D=
{z
2C
:|z|
1}.
Pictures
ofother
regions
thatm
ustcontain
arootoff(x).
Where
did
these
regions
com
efrom
?
Why
are
they
ofinterest
to
me?
What
are
or
are
not
other
possible
regions
ofinterest?
aroot
out
here
f(x)
monic
=)
The
Motivation
(from
my
point
ofview
)
Theorem
(A
.C
ohn).
Let
d
n
d
n�1
...d
1
d
0
be
the
decim
al
representation
ofa
prim
e>
10,and
let
f(x)
=d
n
x
n
+d
n�1
x
n�1
+···
+d
1
x+
d
0
.
Then
f(x)
is
irreducible
over
the
rationals.
Theore
m(A
.C
ohn).
Let
f(x)
be
anon-constant
poly-
nom
ialwith
non-negative
integer
coe�
cients
satisfying
f(1
0)
is
prim
e.
If
the
coe�
cients
are
each
9,
then
f(x)
is
irreducible.
The
Motivation
(from
my
point
ofview
)
Joint
work
with
Morgan
Cole,Scott
Dunn
and
Sam
Gross
Theore
m(A
.C
ohn).
Let
f(x)
be
anon-constant
poly-
nom
ialwith
non-negative
integer
coe�
cients
satisfying
f(1
0)
is
prim
e.
If
the
coe�
cients
are
each
9,
then
f(x)
is
irreducible.
Theorem
.Let
f(x)
be
apolynom
ialhaving
non-negative
integer
coe�
cients
and
satisfying
f(10)
is
prim
e.
Ifthe
coe�
cients
are
each
49598666989151226098104244512918,
then
f(x)
is
irreducible.
Ifinstead
the
coe�
cients
are
8592444743529135815769545955936773
and
f(x)
is
reducible,then
f(x)
is
divisible
by
x
2�20x
+101.
Ifinstead
the
coe�
cients
are
2.749
⇥10
35
and
f(x)
is
reducible,then
f(x)
is
divisible
by
one
of
x
2�20x
+101
and
x
2�19x
+91.
Theorem
.Let
f(x)
be
apolynom
ialhaving
non-negative
integer
coe�
cients
and
satisfying
f(10)
is
prim
e.
Ifthe
coe�
cients
are
each
49598666989151226098104244512918,
then
f(x)
is
irreducible.
Ifinstead
the
coe�
cients
are
8592444743529135815769545955936773
and
f(x)
is
reducible,then
f(x)
is
divisible
by
x
2�20x
+101.
Ifinstead
the
coe�
cients
are
2.749
⇥10
35
and
f(x)
is
reducible,then
f(x)
is
divisible
by
one
of
x
2�20x
+101
and
x
2�19x
+91.
Theorem
.Let
f(x)
be
apolynom
ialhaving
non-negative
integer
coe�
cients
and
satisfying
f(10)
is
prim
e.
Ifthe
coe�
cients
are
each
49598666989151226098104244512918,
then
f(x)
is
irreducible.
Ifinstead
the
coe�
cients
are
8592444743529135815769545955936773
and
f(x)
is
reducible,then
f(x)
is
divisible
by
x
2�20x
+101.
Ifinstead
the
coe�
cients
are
2.749
⇥10
35
and
f(x)
is
reducible,then
f(x)
is
divisible
by
one
of
x
2�20x
+101
and
x
2�19x
+91.
Theore
m(A
.C
ohn).
Let
d
n
d
n�1...d
1d
0be
the
decim
al
representation
ofa
prim
e>
10,and
let
f(x)
=d
n
x
n
+d
n�1x
n�1+
···+
d
1x
+d
0.
Then
f(x)
is
irreducible
over
the
rationals.
One
of
many
libra
ry
books
now
loca
ted
in
my
o�
ceat
LC
301.
Exam
ple
.Sin
ce7776589
isa
prim
e,
7x
6+
7x
5+
7x
4+
6x
3+
5x
2+
8x
+9
isirre
ducib
le.
Basic
Idea:
Iff(x)
=g(x)h(x)w
ithg(x)and
h(x)in
Z[x],
then
f(1
0)
would
have
the
facto
rizatio
ng(1
0)h(1
0).
Basic
Fla
win
Basic
Idea:
What
ifg(1
0)
=1?
Theore
m(A
.C
ohn).
Let
d
n
d
n�1...d
1d
0be
the
decim
al
representation
ofa
prim
e>
10,and
let
f(x)
=d
n
x
n
+d
n�1x
n�1+
···+
d
1x
+d
0.
Then
f(x)
is
irreducible
over
the
rationals.
One
of
many
libra
ry
books
now
loca
ted
in
my
o�
ceat
LC
301.
Exam
ple
.Sin
ce7776589
isa
prim
e,
7x
6+
7x
5+
7x
4+
6x
3+
5x
2+
8x
+9
isirre
ducib
le.
Basic
Idea:
Iff(x)
=g(x)h(x)w
ithg(x)and
h(x)in
Z[x],
then
f(1
0)
would
have
the
facto
rizatio
ng(1
0)h(1
0).
Basic
Fla
win
Basic
Idea:
What
ifg(1
0)
=1?
Theore
m(A
.C
ohn).
Let
d
n
d
n�1...d
1d
0be
the
decim
al
representation
ofa
prim
e>
10,and
let
f(x)
=d
n
x
n
+d
n�1x
n�1+
···+
d
1x
+d
0.
Then
f(x)
is
irreducible
over
the
rationals.
One
of
many
libra
ry
books
now
loca
ted
in
my
o�
ceat
LC
301.
Exam
ple
.Sin
ce7776589
isa
prim
e,
7x
6+
7x
5+
7x
4+
6x
3+
5x
2+
8x
+9
isirre
ducib
le.
Basic
Idea:
Iff(x)
=g(x)h(x)w
ithg(x)and
h(x)in
Z[x],
then
f(1
0)
would
have
the
facto
rizatio
ng(1
0)h(1
0).
Basic
Fla
win
Basic
Idea:
What
ifg(1
0)
=1?
The
Motivation
(from
my
point
ofview
)
Joint
work
with
Morgan
Cole,Scott
Dunn
and
Sam
Gross
Com
ment:
In
thiscase,g(x)
2Z[x]and
g(10)
=1,so
g(x)
hasa
rootin
any
region
asdescribed
earlier
buttranslated
to
the
right
10.
Theore
m(A
.C
ohn).
Let
f(x)
be
anon-constant
poly-
nom
ialwith
non-negative
integer
coe�
cients
satisfying
f(1
0)
is
prim
e.
If
the
coe�
cients
are
each
9,
then
f(x)
is
irreducible.
✓
max
3.
Ido
not
know
ifso
meth
ing
simila
rco
ntin
ues
to
happen
here
.It
may
be
the
case
that
for
som
e
bound
on
the
coe�
cients,
are
ducib
lef(x)
does
not
have
tohave
afa
ctor
inany
specifi
ed
finite
list
ofpoly
nom
ials.
What
can
we
say
about
the
imagin
ary
part
of
f(↵
)if
1
deg
f
31
?
f(x)
=n
X
j=
0
a
j
x
j
reducib
le,
f(1
0)
prim
e
a
n
>0,
a
j �0
for
all
j
↵=
re
i✓
Why
must
the
coe�
cients
of
f(x)
be
larg
e?
����f(↵
)
↵
n
����=
����a
n
+a
n�1↵
�1+
···+
a
n�15↵
�15+
n
Xj=
16
a
n�j
↵
�j ����
�1
�n
Xj=
16
max{|
a
j |}r
j
>1
�m
ax{|
a
j |}r
15(
r�
1)
max{|
a
j |}>
915·
8
What
can
we
say
about
apoly
nom
ial
g(x)
2Z[x]th
at
satisfi
es
g(1
0)
=1?
Answ
er
1:
Ithas
aro
ot
inD
={z
:|z
�10|
1}.
re
i✓
r�
90
<✓
sin
�1(1
/10)
=0.1
001674
...
What
can
we
say
about
apoly
nom
ial
g(x)
2Z[x]th
at
satisfi
es
g(1
0)
=1?
Answ
er
1:
Ithas
aro
ot
inD
={z
:|z
�10|
1}.
re
i✓
r�
90
<✓
<sin
�1(1
/10)
=0.1
001674
...
The
Motivation
(from
my
point
ofview
)
Joint
work
with
Morgan
Cole,Scott
Dunn
and
Sam
Gross
Com
ment:
In
thiscase,g(x)
2Z[x]and
g(10)
=1,so
g(x)
hasa
rootin
any
region
asdescribed
earlier
buttranslated
to
the
right
10.
The
Motivation
(from
my
point
ofview
)
Joint
work
with
Morgan
Cole,Scott
Dunn
and
Sam
Gross
Com
ment:
In
thiscase,g(x)
2Z[x]and
g(10)
=1,so
g(x)
hasa
rootin
any
region
asdescribed
earlier
buttranslated
to
the
right
10.
Initially,consider
the
region
D=
{z
2C
:|z
�10|
1}.
For
k=
32,
the
expression
|↵
|k
is
large
which
forces
the
coe�
cients
to
be
large
(so
that
f(↵
)=
0).
Note
that
for
k>
32,the
expression
|↵
|k
is
even
larger.
f(x)
=n
X
j=
0
a
j
x
j
reducib
le,
f(1
0)
prim
e
a
n
>0,
a
j �0
for
all
j
↵=
re
i✓
Why
must
the
coe�
cients
of
f(x)
be
larg
e?
����f(↵
)
↵
n
����=
����a
n
+a
n�1↵
�1+
···+
a
n�15↵
�15+
n
Xj=
16
a
n�j
↵
�j ����
�1
�n
Xj=
16
max{|
a
j |}r
j
>1
�m
ax{|
a
j |}r
15(
r�
1)
max{|
a
j |}>
915·
8
What
can
we
say
about
apoly
nom
ial
g(x)
2Z[x]th
at
satisfi
es
g(1
0)
=1?
Answ
er
1:
Ithas
aro
ot
inD
={z
:|z
�10|
1}.
re
i✓
r�
90
<✓
sin
�1(1
/10)
=0.1
001674
...
What
can
we
say
about
the
imagin
ary
part
of
↵
k
=(re
i✓)
k
if1
k
31
?
Theore
m.
Let
f(x)
=P
nj=
0a
j
x
j
2Z[x]satisfying
a
j
�0
for
each
jand
f(1
0)
is
aprim
e.
If
1
deg
f
31,then
f(x)
is
irreducible.
Theore
m.
Let
f(x)
be
anon-constant
polynom
ial
with
non-negative
integercoe�
cientssatisfying
f(1
0)isprim
e
and
deg
f
31.
Then
f(x)
is
irreducible.
The
Motivation
(from
my
point
ofview
)
Joint
work
with
Morgan
Cole,Scott
Dunn
and
Sam
Gross
Com
ment:
In
thiscase,g(x)
2Z[x]and
g(10)
=1,so
g(x)
hasa
rootin
any
region
asdescribed
earlier
buttranslated
to
the
right
10.
Theore
m(A
.C
ohn).
Let
f(x)
be
anon-constant
poly-
nom
ialwith
non-negative
integer
coe�
cients
satisfying
f(1
0)
is
prim
e.
If
the
coe�
cients
are
each
9,
then
f(x)
is
irreducible.
✓
max
3.
Ido
not
know
ifso
meth
ing
simila
rco
ntin
ues
to
happen
here
.It
may
be
the
case
that
for
som
e
bound
on
the
coe�
cients,
are
ducib
lef(x)
does
not
have
tohave
afa
ctor
inany
specifi
ed
finite
list
ofpoly
nom
ials.
What
can
we
say
about
the
imagin
ary
part
of
f(↵
)if
1
deg
f
31
?
What
can
we
say
about
apoly
nom
ial
g(x)
2Z[x]th
at
satisfi
es
g(1
0)
=1?
Answ
er
1:
Ithas
aro
ot
inD
={z
:|z
�10|
1}.
re
i✓
r�
90
<✓
sin
�1(1
/10)
=0.1
001674
...
What
can
we
say
about
apoly
nom
ial
g(x)
2Z[x]th
at
satisfi
es
g(1
0)
=1?
Answ
er
1:
Ithas
aro
ot
inD
={z
:|z
�10|
1}.
re
i✓
r�
90
<✓
<sin
�1(1
/10)
=0.1
001674
...
The
Motivation
(from
my
point
ofview
)
Joint
work
with
Morgan
Cole,Scott
Dunn
and
Sam
Gross
Com
ment:
In
thiscase,g(x)
2Z[x]and
g(10)
=1,so
g(x)
hasa
rootin
any
region
asdescribed
earlier
buttranslated
to
the
right
10.
Initially,consider
the
region
D=
{z
2C
:|z
�10|
1}.
For
k=
32,
the
expression
|↵
|k
is
large
which
forces
the
coe�
cients
to
be
large
(so
that
f(↵
)=
0).
Note
that
for
k>
32,the
expression
|↵
|k
is
even
larger.
f(x)
=n
X
j=
0
a
j
x
j
reducib
le,
f(1
0)
prim
e
a
n
>0,
a
j �0
for
all
j
↵=
re
i✓
Why
must
the
coe�
cients
of
f(x)
be
larg
e?
����f(↵
)
↵
n
����=
����a
n
+a
n�1↵
�1+
···+
a
n�15↵
�15+
n
Xj=
16
a
n�j
↵
�j ����
�1
�n
Xj=
16
max{|
a
j |}r
j
>1
�m
ax{|
a
j |}r
15(
r�
1)
max{|
a
j |}>
915·
8
Theore
m(A
.C
ohn).
Let
f(x)
be
anon-constant
poly-
nom
ialwith
non-negative
integer
coe�
cients
satisfying
f(1
0)
is
prim
e.
If
the
coe�
cients
are
each
9,
then
f(x)
is
irreducible.
3.
Ido
not
know
ifso
meth
ing
simila
rco
ntin
ues
to
happen
here
.It
may
be
the
case
that
for
som
e
bound
on
the
coe�
cients,
are
ducib
lef(x)
does
not
have
tohave
afa
ctor
inany
specifi
ed
finite
list
ofpoly
nom
ials.
What
can
we
say
about
the
imagin
ary
part
of
f(↵
)if
1
deg
f
31
?
What
can
we
say
about
apoly
nom
ial
g(x)
2Z[x]th
at
satisfi
es
g(1
0)
=1?
Answ
er
1:
Ithas
aro
ot
inD
={z
:|z
�10|
1}.
re
i✓
r�
90
<✓
sin
�1(1
/10)
=0.1
001674
...
What
can
we
say
about
the
imagin
ary
part
of
↵
k
=(re
i✓)
k
if1
k
31
?
Theore
m.
Let
f(x)
=P
nj=
0a
j
x
j
2Z[x]satisfying
a
j
�0
for
each
jand
f(1
0)
is
aprim
e.
If
1
deg
f
31,then
f(x)
is
irreducible.
Theore
m.
Let
f(x)
be
anon-constant
polynom
ial
with
non-negative
integercoe�
cientssatisfying
f(1
0)isprim
e
and
deg
f
31.
Then
f(x)
is
irreducible.
The
Motivation
(from
my
point
ofview
)
Joint
work
with
Morgan
Cole,Scott
Dunn
and
Sam
Gross
Com
ment:
In
thiscase,g(x)
2Z[x]and
g(10)
=1,so
g(x)
hasa
rootin
any
region
asdescribed
earlier
buttranslated
to
the
right
10.
Theore
m(A
.C
ohn).
Let
f(x)
be
anon-constant
poly-
nom
ialwith
non-negative
integer
coe�
cients
satisfying
f(1
0)
is
prim
e.
If
the
coe�
cients
are
each
9,
then
f(x)
is
irreducible.
✓
max
3.
Ido
not
know
ifso
meth
ing
simila
rco
ntin
ues
to
happen
here
.It
may
be
the
case
that
for
som
e
bound
on
the
coe�
cients,
are
ducib
lef(x)
does
not
have
tohave
afa
ctor
inany
specifi
ed
finite
list
ofpoly
nom
ials.
What
can
we
say
about
the
imagin
ary
part
of
f(↵
)if
1
deg
f
31
?
The
Motivation
(from
my
point
ofview
)
Joint
work
with
Morgan
Cole,Scott
Dunn
and
Sam
Gross
Com
ment:
In
thiscase,g(x)
2Z[x]and
g(10)
=1,so
g(x)
hasa
rootin
any
region
asdescribed
earlier
buttranslated
to
the
right
10.
Initially,consider
the
region
D=
{z
2C
:|z
�10|
1}.
For
k=
32,
the
expression
|↵
|k
is
large
which
forces
the
coe�
cients
to
be
large
(so
that
f(↵
)=
0).
Note
that
for
k>
32,the
expression
|↵
|k
is
even
larger.
What
can
we
say
about
apoly
nom
ial
g(x)
2Z[x]th
at
satisfi
es
g(1
0)
=1?
Answ
er
1:
Ithas
aro
ot
inD
={z
:|z
�10|
1}.
re
i✓
r�
90
<✓
sin
�1(1
/10)
=0.1
001674
...
What
can
we
say
about
the
imagin
ary
part
of(re
i✓)
k
if1
k
31
?
Theore
m.
Let
f(x)
=P
nj=
0a
j
x
j
2Z[x]satisfying
a
j
�0
for
each
jand
f(1
0)
is
aprim
e.
If
1
deg
f
31,then
f(x)
is
irreducible.
Theore
m.
Let
f(x)
be
anon-constant
polynom
ial
with
non-negative
integercoe�
cientssatisfying
f(1
0)isprim
e
and
deg
f
31.
Then
f(x)
is
irreducible.
Theore
m(A
.C
ohn).
Let
f(x)
be
anon-constant
poly-
nom
ialwith
non-negative
integer
coe�
cients
satisfying
f(1
0)
is
prim
e.
If
the
coe�
cients
are
each
9,
then
f(x)
is
irreducible.
3.
Ido
not
know
ifso
meth
ing
simila
rco
ntin
ues
to
happen
here
.It
may
be
the
case
that
for
som
e
bound
on
the
coe�
cients,
are
ducib
lef(x)
does
not
have
tohave
afa
ctor
inany
specifi
ed
finite
list
ofpoly
nom
ials.
What
can
we
say
about
the
imagin
ary
part
of
f(↵
)if
1
deg
f
31
?
What
can
we
say
about
apoly
nom
ial
g(x)
2Z[x]th
at
satisfi
es
g(1
0)
=1?
Answ
er
1:
Ithas
aro
ot
inD
={z
:|z
�10|
1}.
re
i✓
r�
90
<✓
sin
�1(1
/10)
=0.1
001674
...
What
can
we
say
about
the
imagin
ary
part
of
↵
k
=(re
i✓)
k
if1
k
31
?
Theore
m.
Let
f(x)
=P
nj=
0a
j
x
j
2Z[x]satisfying
a
j
�0
for
each
jand
f(1
0)
is
aprim
e.
If
1
deg
f
31,then
f(x)
is
irreducible.
Theore
m.
Let
f(x)
be
anon-constant
polynom
ial
with
non-negative
integercoe�
cientssatisfying
f(1
0)isprim
e
and
deg
f
31.
Then
f(x)
is
irreducible.
The
Motivation
(from
my
point
ofview
)
Joint
work
with
Morgan
Cole,Scott
Dunn
and
Sam
Gross
Com
ment:
In
thiscase,g(x)
2Z[x]and
g(10)
=1,so
g(x)
hasa
rootin
any
region
asdescribed
earlier
buttranslated
to
the
right
10.
The
Motivation
(from
my
point
ofview
)
Joint
work
with
Morgan
Cole,Scott
Dunn
and
Sam
Gross
Com
ment:
In
thiscase,g(x)
2Z[x]and
g(10)
=1,so
g(x)
hasa
rootin
any
region
asdescribed
earlier
buttranslated
to
the
right
10.
Initially,consider
the
region
D=
{z
2C
:|z
�10|
1}.
For
k=
32,
the
expression
|↵
|k
is
large
which
forces
the
coe�
cients
to
be
large
(so
that
f(↵
)=
0).
Note
that
for
k>
32,the
expression
|↵
|k
is
even
larger.
f(x)
=n
X
j=
0
a
j
x
j
reducib
le,
f(1
0)
prim
e
a
n
>0,
a
j �0
for
all
j
↵=
re
i✓
Why
must
the
coe�
cients
of
f(x)
be
larg
e?
����f(↵
)
↵
n
����=
����a
n
+a
n�1↵
�1+
···+
a
n�15↵
�15+
n
Xj=
16
a
n�j
↵
�j ����
�1
�n
Xj=
16
max{|
a
j |}r
j
>1
�m
ax{|
a
j |}r
15(
r�
1)
max{|
a
j |}>
915·
8
What
can
we
say
about
apoly
nom
ial
g(x)
2Z[x]th
at
satisfi
es
g(1
0)
=1?
Answ
er
1:
Ithas
aro
ot
inD
={z
:|z
�10|
1}.
re
i✓
r�
90
<✓
sin
�1(1
/10)
=0.1
001674
...
What
can
we
say
about
apoly
nom
ial
g(x)
2Z[x]th
at
satisfi
es
g(1
0)
=1?
Answ
er
1:
Ithas
aro
ot
inD
={z
:|z
�10|
1}.
re
i✓
r�
90
<✓
<sin
�1(1
/10)
=0.1
001674
...
Theore
m(A
.C
ohn).
Let
f(x)
be
anon-constant
poly-
nom
ialwith
non-negative
integer
coe�
cients
satisfying
f(1
0)
is
prim
e.
If
the
coe�
cients
are
each
9,
then
f(x)
is
irreducible.
✓
max
3.
Ido
not
know
ifso
meth
ing
simila
rco
ntin
ues
to
happen
here
.It
may
be
the
case
that
for
som
e
bound
on
the
coe�
cients,
are
ducib
lef(x)
does
not
have
tohave
afa
ctor
inany
specifi
ed
finite
list
ofpoly
nom
ials.
What
can
we
say
about
the
imagin
ary
part
of
f(↵
)if
1
deg
f
31
?
The
Motivation
(from
my
point
ofview
)
Joint
work
with
Morgan
Cole,Scott
Dunn
and
Sam
Gross
Com
ment:
In
thiscase,g(x)
2Z[x]and
g(10)
=1,so
g(x)
hasa
rootin
any
region
asdescribed
earlier
buttranslated
to
the
right
10.
The
Motivation
(from
my
point
ofview
)
Joint
work
with
Morgan
Cole,Scott
Dunn
and
Sam
Gross
Com
ment:
In
thiscase,g(x)
2Z[x]and
g(10)
=1,so
g(x)
hasa
rootin
any
region
asdescribed
earlier
buttranslated
to
the
right
10.
Initially,consider
the
region
D=
{z
2C
:|z|
1}.
For
k=
32,
the
expression
|↵
|k
is
large
which
forces
the
coe�
cients
to
be
large
(so
that
f(↵
)=
0).
Note
that
for
k>
32,the
expression
|↵
|k
is
even
larger.
The
Motivation
(from
my
point
ofview
)
Joint
work
with
Morgan
Cole,Scott
Dunn
and
Sam
Gross
Com
ment:
In
thiscase,g(x)
2Z[x]and
g(10)
=1,so
g(x)
hasa
rootin
any
region
asdescribed
earlier
buttranslated
to
the
right
10.
Initially,consider
the
region
D=
{z
2C
:|z|
1}.
For
k=
32,
the
expression
|↵
|k
is
large
which
forces
the
coe�
cients
to
be
large
(so
that
f(↵
)=
0).
Note
that
for
k>
32,the
expression
|↵
|k
is
even
larger.
The
Motivation
(from
my
point
ofview
)
Joint
work
with
Morgan
Cole,Scott
Dunn
and
Sam
Gross
Com
ment:
In
thiscase,g(x)
2Z[x]and
g(10)
=1,so
g(x)
hasa
rootin
any
region
asdescribed
earlier
buttranslated
to
the
right
10.
Initially,consider
the
region
D=
{z
2C
:|z
�10|
1}.
For
k=
32,
the
expression
|↵
|k
is
large
which
forces
the
coe�
cients
to
be
large
(so
that
f(↵
)=
0).
Note
that
for
k>
32,the
expression
|↵
|k
is
even
larger.
What
can
we
say
about
apoly
nom
ial
g(x)
2Z[x]th
at
satisfi
es
g(1
0)
=1?
Answ
er
1:
Ithas
aro
ot
inD
={z
:|z
�10|
1}.
re
i✓
r�
90
<✓
sin
�1(1
/10)
=0.1
001674
...
The
Motivation
(from
my
point
ofview
)
Joint
work
with
Morgan
Cole,Scott
Dunn
and
Sam
Gross
Com
ment:
In
thiscase,g(x)
2Z[x]and
g(10)
=1,so
g(x)
hasa
rootin
any
region
asdescribed
earlier
buttranslated
to
the
right
10.
The
Motivation
(from
my
point
ofview
)
Joint
work
with
Morgan
Cole,Scott
Dunn
and
Sam
Gross
Com
ment:
In
thiscase,g(x)
2Z[x]and
g(10)
=1,so
g(x)
hasa
rootin
any
region
asdescribed
earlier
buttranslated
to
the
right
10.
Initially,consider
the
region
D=
{z
2C
:|z|
1}.
For
k=
32,
the
expression
|↵
|k
is
large
which
forces
the
coe�
cients
to
be
large
(so
that
f(↵
)=
0).
Note
that
for
k>
32,the
expression
|↵
|k
is
even
larger.
The
Motivation
(from
my
point
ofview
)
Joint
work
with
Morgan
Cole,Scott
Dunn
and
Sam
Gross
Com
ment:
In
thiscase,g(x)
2Z[x]and
g(10)
=1,so
g(x)
hasa
rootin
any
region
asdescribed
earlier
buttranslated
to
the
right
10.
Initially,consider
the
region
D=
{z
2C
:|z|
1}.
For
k=
32,
the
expression
|↵
|k
is
large
which
forces
the
coe�
cients
to
be
large
(so
that
f(↵
)=
0).
Note
that
for
k>
32,the
expression
|↵
|k
is
even
larger.
What
can
we
say
about
apoly
nom
ial
g(x)
2Z[x]th
at
satisfi
es
g(1
0)
=1?
Answ
er
1:
Ithas
aro
ot
inD
={z
:|z
�10|
1}.
re
i✓
r�
90
<✓
<sin
�1(1
/10)
=0.1
001674
...
The
Motivation
(from
my
point
ofview
)
Joint
work
with
Morgan
Cole,Scott
Dunn
and
Sam
Gross
Com
ment:
In
thiscase,g(x)
2Z[x]and
g(10)
=1,so
g(x)
hasa
rootin
any
region
asdescribed
earlier
buttranslated
to
the
right
10.
Initially,consider
the
region
D=
{z
2C
:|z
�10|
1}.
For
k=
32,
the
expression
|↵
|k
is
large
which
forces
the
coe�
cients
to
be
large
(so
that
f(↵
)=
0).
Note
that
for
k>
32,the
expression
|↵
|k
is
even
larger.
Theore
m(A
.C
ohn).
Let
f(x)
be
anon-constant
poly-
nom
ialwith
non-negative
integer
coe�
cients
satisfying
f(1
0)
is
prim
e.
If
the
coe�
cients
are
each
9,
then
f(x)
is
irreducible.
✓
max
3.
Ido
not
know
ifso
meth
ing
simila
rcontin
ues
to
happen
here
.It
may
be
the
case
that
for
som
e
bound
on
the
coe�
cie
nts,
are
ducib
lef(x)
does
not
have
tohave
afa
cto
rin
any
specifi
ed
finite
list
ofpoly
nom
ials.
What
can
we
say
about
the
imagin
ary
part
of
f(↵
)if
1
deg
f
31
?
f(x)
=n
X
j=
0
a
j
x
j
reducib
le,
f(1
0)
prim
e
a
n
>0,
a
j �0
for
all
j
↵=
re
i✓
Why
must
the
coe�
cients
of
f(x)
be
larg
e?
����f(↵
)
↵
n
����=
����a
n
+a
n�1↵
�1+
···+
a
n�15↵
�15+
n
Xj=
16
a
n�j
↵
�j ����
�1
�n
Xj=
16
max{|
a
j |}r
j
>1
�m
ax{|
a
j |}r
15(
r�
1)
max{|
a
j |}>
915·
8
What
can
we
say
about
apoly
nom
ial
g(x)
2Z[x]th
at
satisfi
es
g(1
0)
=1?
Answ
er
1:
Ithas
aro
ot
inD
={z
:|z
�10|
1}.
re
i✓
r�
90
<✓
sin
�1(1
/10)
=0.1
001674
...
The
Motivation
(from
my
point
ofview
)
Joint
work
with
Morgan
Cole,Scott
Dunn
and
Sam
Gross
Com
ment:
In
thiscase,g(x)
2Z[x]and
g(10)
=1,so
g(x)
hasa
rootin
any
region
asdescribed
earlier
buttranslated
to
the
right
10.
The
Motivation
(from
my
point
ofview
)
Joint
work
with
Morgan
Cole,Scott
Dunn
and
Sam
Gross
Com
ment:
In
thiscase,g(x)
2Z[x]and
g(10)
=1,so
g(x)
hasa
rootin
any
region
asdescribed
earlier
buttranslated
to
the
right
10.
Initially,consider
the
region
D=
{z
2C
:|z|
1}.
For
k=
32,
the
expression
|↵
|k
is
large
which
forces
the
coe�
cients
to
be
large
(so
that
f(↵
)=
0).
Note
that
for
k>
32,the
expression
|↵
|k
is
even
larger.
The
Motivation
(from
my
point
ofview
)
Joint
work
with
Morgan
Cole,Scott
Dunn
and
Sam
Gross
Com
ment:
In
thiscase,g(x)
2Z[x]and
g(10)
=1,so
g(x)
hasa
rootin
any
region
asdescribed
earlier
buttranslated
to
the
right
10.
Initially,consider
the
region
D=
{z
2C
:|z|
1}.
For
k=
32,
the
expression
|↵
|k
is
large
which
forces
the
coe�
cients
to
be
large
(so
that
f(↵
)=
0).
Note
that
for
k>
32,the
expression
|↵
|k
is
even
larger.
What
can
we
say
about
apoly
nom
ial
g(x)
2Z[x]th
at
satisfi
es
g(1
0)
=1?
Answ
er
1:
Ithas
aro
ot
inD
={z
:|z
�10|
1}.
re
i✓
r�
90
<✓
<sin
�1(1
/10)
=0.1
001674
...
The
Motivation
(from
my
point
ofview
)
Joint
work
with
Morgan
Cole,Scott
Dunn
and
Sam
Gross
Com
ment:
In
thiscase,g(x)
2Z[x]and
g(10)
=1,so
g(x)
hasa
rootin
any
region
asdescribed
earlier
buttranslated
to
the
right
10.
Initially,consider
the
region
D=
{z
2C
:|z
�10|
1}.
For
k=
32,
the
expression
|↵
|k
is
large
which
forces
the
coe�
cients
to
be
large
(so
that
f(↵
)=
0).
Note
that
for
k>
32,the
expression
|↵
|k
is
even
larger.
0i
�1+
p3
i
2
1+
p3
i
2
✓<
⇡/36
The
smalle
r✓
is,th
ela
rger
one
can
take
k.
Answ
er
2:
Ithas
aro
ot
inth
ere
gio
nco
lore
dbelo
w.
Lem
ma.
If
g(x)
2Z[x]satisfies
g(1
0)
=1,
then
either
g(x)
is
divisible
by
one
of
(x
�10)2+
1=
x
2�20x
+101,
(x
�10)2+
(x
�10)+
1=
x
2�19x
+91,
and
(x
�10)2�
(x
�10)+
1=
x
2�21x
+111
or
g(x)
has
arootin
the
“region”
Rbelow.
The
Motivation
(from
my
point
ofview
)
Joint
work
with
Morgan
Cole,Scott
Dunn
and
Sam
Gross
Com
ment:
In
thiscase,g(x)
2Z[x]and
g(10)
=1,so
g(x)
hasa
rootin
any
region
asdescribed
earlier
buttranslated
to
the
right
10.
The
Motivation
(from
my
point
ofview
)
Joint
work
with
Morgan
Cole,Scott
Dunn
and
Sam
Gross
Com
ment:
In
thiscase,g(x)
2Z[x]and
g(10)
=1,so
g(x)
hasa
rootin
any
region
asdescribed
earlier
buttranslated
to
the
right
10.
Initially,consider
the
region
D=
{z
2C
:|z|
1}.
For
k=
32,
the
expression
|↵
|k
is
large
which
forces
the
coe�
cients
to
be
large
(so
that
f(↵
)=
0).
Note
that
for
k>
32,the
expression
|↵
|k
is
even
larger.
The
Motivation
(from
my
point
ofview
)
Joint
work
with
Morgan
Cole,Scott
Dunn
and
Sam
Gross
Com
ment:
In
thiscase,g(x)
2Z[x]and
g(10)
=1,so
g(x)
hasa
rootin
any
region
asdescribed
earlier
buttranslated
to
the
right
10.
Initially,consider
the
region
D=
{z
2C
:|z|
1}.
For
k=
32,
the
expression
|↵
|k
is
large
which
forces
the
coe�
cients
to
be
large
(so
that
f(↵
)=
0).
Note
that
for
k>
32,the
expression
|↵
|k
is
even
larger.
What
can
we
say
about
apoly
nom
ial
g(x)
2Z[x]th
at
satisfi
es
g(1
0)
=1?
Answ
er
1:
Ithas
aro
ot
inD
={z
:|z
�10|
1}.
re
i✓
r�
90
<✓
<sin
�1(1
/10)
=0.1
001674
...
Theore
m(A
.C
ohn).
Let
f(x)
be
anon-constant
poly-
nom
ialwith
non-negative
integer
coe�
cients
satisfying
f(1
0)
is
prim
e.
If
the
coe�
cients
are
each
9,
then
f(x)
is
irreducible.
✓
max
3.
Ido
not
know
ifso
meth
ing
simila
rcontin
ues
to
happen
here
.It
may
be
the
case
that
for
som
e
bound
on
the
coe�
cie
nts,
are
ducib
lef(x)
does
not
have
tohave
afa
cto
rin
any
specifi
ed
finite
list
ofpoly
nom
ials.
What
can
we
say
about
the
imagin
ary
part
of
f(↵
)if
1
deg
f
31
?
f(x)
=n
X
j=
0
a
j
x
j
reducib
le,
f(1
0)
prim
e
a
n
>0,
a
j �0
for
all
j
↵=
re
i✓
Why
must
the
coe�
cients
of
f(x)
be
larg
e?
����f(↵
)
↵
n
����=
����a
n
+a
n�1↵
�1+
···+
a
n�15↵
�15+
n
Xj=
16
a
n�j
↵
�j ����
�1
�n
Xj=
16
max{|
a
j |}r
j
>1
�m
ax{|
a
j |}r
15(
r�
1)
max{|
a
j |}>
915·
8
The
Motivation
(from
my
point
ofview
)
Joint
work
with
Morgan
Cole,Scott
Dunn
and
Sam
Gross
Com
ment:
In
thiscase,g(x)
2Z[x]and
g(10)
=1,so
g(x)
hasa
rootin
any
region
asdescribed
earlier
buttranslated
to
the
right
10.
Initially,consider
the
region
D=
{z
2C
:|z
�10|
1}.
For
k=
32,
the
expression
|↵
|k
is
large
which
forces
the
coe�
cients
to
be
large
(so
that
f(↵
)=
0).
Note
that
for
k>
32,the
expression
|↵
|k
is
even
larger.
0i
�1+
p3
i
2
1+
p3
i
2
✓<
⇡/36
=)k
�37
(for
↵2
R)
The
smalle
r✓
is,th
ela
rger
one
can
take
k.
Answ
er
2:
Ithas
aro
ot
inth
ere
gio
nco
lore
dbelo
w.
Lem
ma.
If
g(x)
2Z[x]satisfies
g(1
0)
=1,
then
either
g(x)
is
divisible
by
one
of
(x
�10)2+
1=
x
2�20x
+101,
(x
�10)2+
(x
�10)+
1=
x
2�19x
+91,
and
(x
�10)2�
(x
�10)+
1=
x
2�21x
+111
or
g(x)
has
arootin
the
“region”
Rbelow.
Theorem
.Let
f(x)
be
apolynom
ialhaving
non-negative
integer
coe�
cients
and
satisfying
f(10)
is
prim
e.
Ifthe
coe�
cients
are
each
49598666989151226098104244512918,
then
f(x)
is
irreducible.
Ifinstead
the
coe�
cients
are
8592444743529135815769545955936773
and
f(x)
is
reducible,then
f(x)
is
divisible
by
x
2�20x
+101.
Ifinstead
the
coe�
cients
are
2.749
⇥10
35
and
f(x)
is
reducible,then
f(x)
is
divisible
by
one
of
x
2�20x
+101
and
x
2�19x
+91.
0i
�1+
p3
i
2
1+
p3
i
2
✓<
⇡/36
The
smalle
r✓
is,th
ela
rger
one
can
take
k.
Answ
er
2:
Ithas
aro
ot
inth
ere
gio
nco
lore
dbelo
w.
Lem
ma.
If
g(x)
2Z[x]satisfies
g(1
0)
=1,
then
either
g(x)
is
divisible
by
one
of
(x
�10)2+
1=
x
2�20x
+101,
(x
�10)2+
(x
�10)+
1=
x
2�19x
+91,
and
(x
�10)2�
(x
�10)+
1=
x
2�21x
+111
or
g(x)
has
arootin
the
“region”
Rbelow.
The
Motivation
(from
my
point
ofview
)
Joint
work
with
Morgan
Cole,Scott
Dunn
and
Sam
Gross
Com
ment:
In
thiscase,g(x)
2Z[x]and
g(10)
=1,so
g(x)
hasa
rootin
any
region
asdescribed
earlier
buttranslated
to
the
right
10.
The
Motivation
(from
my
point
ofview
)
Joint
work
with
Morgan
Cole,Scott
Dunn
and
Sam
Gross
Com
ment:
In
thiscase,g(x)
2Z[x]and
g(10)
=1,so
g(x)
hasa
rootin
any
region
asdescribed
earlier
buttranslated
to
the
right
10.
Initially,consider
the
region
D=
{z
2C
:|z|
1}.
For
k=
32,
the
expression
|↵
|k
is
large
which
forces
the
coe�
cients
to
be
large
(so
that
f(↵
)=
0).
Note
that
for
k>
32,the
expression
|↵
|k
is
even
larger.
The
Motivation
(from
my
point
ofview
)
Joint
work
with
Morgan
Cole,Scott
Dunn
and
Sam
Gross
Com
ment:
In
thiscase,g(x)
2Z[x]and
g(10)
=1,so
g(x)
hasa
rootin
any
region
asdescribed
earlier
buttranslated
to
the
right
10.
Initially,consider
the
region
D=
{z
2C
:|z|
1}.
For
k=
32,
the
expression
|↵
|k
is
large
which
forces
the
coe�
cients
to
be
large
(so
that
f(↵
)=
0).
Note
that
for
k>
32,the
expression
|↵
|k
is
even
larger.
What
can
we
say
about
apoly
nom
ial
g(x)
2Z[x]th
at
satisfi
es
g(1
0)
=1?
Answ
er
1:
Ithas
aro
ot
inD
={z
:|z
�10|
1}.
re
i✓
r�
90
<✓
<sin
�1(1
/10)
=0.1
001674
...
Theore
m(A
.C
ohn).
Let
f(x)
be
anon-constant
poly-
nom
ialwith
non-negative
integer
coe�
cients
satisfying
f(1
0)
is
prim
e.
If
the
coe�
cients
are
each
9,
then
f(x)
is
irreducible.
✓
max
3.
Ido
not
know
ifso
meth
ing
simila
rcontin
ues
to
happen
here
.It
may
be
the
case
that
for
som
e
bound
on
the
coe�
cie
nts,
are
ducib
lef(x)
does
not
have
tohave
afa
cto
rin
any
specifi
ed
finite
list
ofpoly
nom
ials.
What
can
we
say
about
the
imagin
ary
part
of
f(↵
)if
1
deg
f
31
?
f(x)
=n
X
j=
0
a
j
x
j
reducib
le,
f(1
0)
prim
e
a
n
>0,
a
j �0
for
all
j
↵=
re
i✓
Why
must
the
coe�
cients
of
f(x)
be
larg
e?
����f(↵
)
↵
n
����=
����a
n
+a
n�1↵
�1+
···+
a
n�15↵
�15+
n
Xj=
16
a
n�j
↵
�j ����
�1
�n
Xj=
16
max{|
a
j |}r
j
>1
�m
ax{|
a
j |}r
15(
r�
1)
max{|
a
j |}>
915·
8
The
Motivation
(from
my
point
ofview
)
Joint
work
with
Morgan
Cole,Scott
Dunn
and
Sam
Gross
Com
ment:
In
thiscase,g(x)
2Z[x]and
g(10)
=1,so
g(x)
hasa
rootin
any
region
asdescribed
earlier
buttranslated
to
the
right
10.
Initially,consider
the
region
D=
{z
2C
:|z
�10|
1}.
For
k=
32,
the
expression
|↵
|k
is
large
which
forces
the
coe�
cients
to
be
large
(so
that
f(↵
)=
0).
Note
that
for
k>
32,the
expression
|↵
|k
is
even
larger.
0i
�1+
p3
i
2
1+
p3
i
2
✓<
⇡/36
=)k
�37
(for
↵2
R)
The
smalle
r✓
is,th
ela
rger
one
can
take
k.
Answ
er
2:
Ithas
aro
ot
inth
ere
gio
nco
lore
dbelo
w.
Lem
ma.
If
g(x)
2Z[x]satisfies
g(1
0)
=1,
then
either
g(x)
is
divisible
by
one
of
(x
�10)2+
1=
x
2�20x
+101,
(x
�10)2+
(x
�10)+
1=
x
2�19x
+91,
and
(x
�10)2�
(x
�10)+
1=
x
2�21x
+111
or
g(x)
has
arootin
the
“region”
Rbelow.
Theorem
.Let
f(x)
be
apolynom
ialhaving
non-negative
integer
coe�
cients
and
satisfying
f(10)
is
prim
e.
Ifthe
coe�
cients
are
each
49598666989151226098104244512918,
then
f(x)
is
irreducible.
Ifinstead
the
coe�
cients
are
8592444743529135815769545955936773
and
f(x)
is
reducible,then
f(x)
is
divisible
by
x
2�20x
+101.
Ifinstead
the
coe�
cients
are
2.749
⇥10
35
and
f(x)
is
reducible,then
f(x)
is
divisible
by
one
of
x
2�20x
+101
and
x
2�19x
+91.
Lem
ma.
Let
f(x)
2Z[x]have
non-negative
coe�
cients
each
49598666989151226098104244512917.
Then
f(x)
is
notdivisible
by
ag(x)
=(x
�10)
2
+1
=x
2�20x
+101.
Lem
ma.
Let
f(x)
2Z[x]have
non-negative
coe�
cients
each
49598666989151226098104244512917.
Then
f(x)
is
notdivisible
by
ag(x)
2Z[x]satisfying
deg
g�
1and
g(10)
=1.
Lem
ma.
Let
f(x)
2Z[x]have
non-negative
coe�
cients
each
49598666989151226098104244512917.
Then
f(x)
is
notdivisible
by
ag(x)
=(x
�10)
2
+1
=x
2�20x
+101.
Lem
ma.
Let
f(x)
2Z[x]have
non-negative
coe�
cients
each
49598666989151226098104244512917.
Then
f(x)
is
notdivisible
by
ag(x)
2Z[x]satisfying
deg
g�
1and
g(10)
=1.
Lem
ma.
Let
f(x)
2Z[x]have
non-negative
coe�
cients
each
49598666989151226098104244512917.
Then
f(x)
is
notdivisible
by
ag(x)
=(x
�10)
2
+1
=x
2�20x
+101.
Lem
ma.
Let
f(x)
2Z[x]have
non-negative
coe�
cients
each
49598666989151226098104244512917.
Then
f(x)
is
notdivisible
by
ag(x)
2Z[x]satisfying
deg
g�
1and
g(10)
=1.
Question
To
what
extent
can
this
region
Rbe
modified?
Dobrow
olskihasprovided
me
with
an
argum
entthatthere
is
not
asim
ilar
region
lying
below
the
line
Im
(z)
<1/2.
Lem
ma.
Let
f(x)
2Z[x]have
non-negative
coe�
cients
each
49598666989151226098104244512917.
Then
f(x)
is
notdivisible
by
ag(x)
=(x
�10)
2
+1
=x
2�20x
+101.
Lem
ma.
Let
f(x)
2Z[x]have
non-negative
coe�
cients
each
49598666989151226098104244512917.
Then
f(x)
is
notdivisible
by
ag(x)
2Z[x]satisfying
deg
g�
1and
g(10)
=1.
Question
To
what
extent
can
these
regions
be
modified?
Dobrow
olskihasprovided
me
with
an
argum
entthatthere
is
not
asim
ilar
region
lying
below
the
line
Im
(z)
<1/2.
Lem
ma.
Let
f(x)
2Z[x]have
non-negative
coe�
cients
each
49598666989151226098104244512917.
Then
f(x)
is
notdivisible
by
ag(x)
=(x
�10)
2
+1
=x
2�20x
+101.
Lem
ma.
Let
f(x)
2Z[x]have
non-negative
coe�
cients
each
49598666989151226098104244512917.
Then
f(x)
is
notdivisible
by
ag(x)
2Z[x]satisfying
deg
g�
1and
g(10)
=1.
Question
To
what
extent
can
this
region
Rbe
modified?
Dobrow
olskihasprovided
me
with
an
argum
entthatthere
is
not
asim
ilar
region
lying
below
the
line
Im
(z)
<1/2.
Lem
ma.
Let
f(x)
2Z[x]have
non-negative
coe�
cients
each
49598666989151226098104244512917.
Then
f(x)
is
notdivisible
by
ag(x)
=(x
�10)
2
+1
=x
2�20x
+101.
Lem
ma.
Let
f(x)
2Z[x]have
non-negative
coe�
cients
each
49598666989151226098104244512917.
Then
f(x)
is
notdivisible
by
ag(x)
2Z[x]satisfying
deg
g�
1and
g(10)
=1.
Question
To
what
extent
can
these
regions
be
modified?
Dobrow
olskihasprovided
me
with
an
argum
entthatthere
is
not
asim
ilar
region
lying
below
the
line
Im
(z)
<1/2.
Lem
ma.
Let
f(x)
2Z[x]have
non-negative
coe�
cients
each
49598666989151226098104244512917.
Then
f(x)
is
notdivisible
by
ag(x)
=(x
�10)
2
+1
=x
2�20x
+101.
Lem
ma.
Let
f(x)
2Z[x]have
non-negative
coe�
cients
each
49598666989151226098104244512917.
Then
f(x)
is
notdivisible
by
ag(x)
2Z[x]satisfying
deg
g�
1and
g(10)
=1.
Question
To
what
extent
can
these
regions
be
modified?
Dobrow
olskihasprovided
me
with
an
argum
entthatthere
is
not
asim
ilar
region
lying
below
the
line
Im
(z)
<1/2.
Lem
ma.
Let
f(x)
2Z[x]have
non-negative
coe�
cients
each
49598666989151226098104244512917.
Then
f(x)
is
notdivisible
by
ag(x)
=(x
�10)
2
+1
=x
2�20x
+101.
Lem
ma.
Let
f(x)
2Z[x]have
non-negative
coe�
cients
each
49598666989151226098104244512917.
Then
f(x)
is
notdivisible
by
ag(x)
2Z[x]satisfying
deg
g�
1and
g(10)
=1.
Question
To
what
extent
can
this
region
Rbe
modified?
Dobrow
olskihasprovided
me
with
an
argum
entthatthere
is
not
asim
ilar
region
lying
below
the
line
Im
(z)
<1/2.
Lem
ma.
Let
f(x)
2Z[x]have
non-negative
coe�
cients
each
49598666989151226098104244512917.
Then
f(x)
is
notdivisible
by
ag(x)
=(x
�10)
2
+1
=x
2�20x
+101.
Lem
ma.
Let
f(x)
2Z[x]have
non-negative
coe�
cients
each
49598666989151226098104244512917.
Then
f(x)
is
notdivisible
by
ag(x)
2Z[x]satisfying
deg
g�
1and
g(10)
=1.
Question
To
what
extent
can
these
regions
be
modified?
Dobrow
olskihasprovided
me
with
an
argum
entthatthere
is
not
asim
ilar
region
lying
below
the
line
Im
(z)
<1/2.
Theorem
.Let
U(b)be
indica
tedbe
low.
Iff(x)
2Z[x]has
non-n
egativ
eco
e�
cie
nts
bounded
by
U(b)
and
issu
ch
that
f(b)
isprim
e,th
en
f(x)
isirred
ucib
le.
bU
(b)
27
33795
48925840
556446139763
6568059199631352
74114789794835622912
875005556404194608192050
91744054672674891153663590400
10
49598666989151226098104244512918
bU
(b)
11
1754638089240473418053140582402752512
12
77040233750234318697380885880167588145722
13
4163976197614743889240641877839816882986680320
14
274327682731486702351640132483696971555362645663790
15
53237820409607236753887375170676537338756637987992240128
16
8267439025097901738248191414518610393726802935783728327213632
17
1268514052720791756582944613802085175096200858994963359873275789312
18
210075378544004872190325829606836051632192371202216081668284609637499040
19
38625368655808052927694359301620272576822252200247254369696128549408630374400
20
7965097815841643900684276577174036821605756035173863133380627982979718588470528880
Theorem
.Let
U(b)be
indica
tedbe
low.
Iff(x)
2Z[x]has
non-n
egativ
eco
e�
cie
nts
bounded
by
U(b)
and
issu
ch
that
f(b)
isprim
e,th
en
f(x)
isirred
ucib
le.
bU
(b)
27
33795
48925840
556446139763
6568059199631352
74114789794835622912
875005556404194608192050
91744054672674891153663590400
10
49598666989151226098104244512918
The End