We prove part (1) using an argument by contradiction.
Thus, suppose there does exist a counterexample to the claim, i.e., a nonzero polynomial pp of degree nn and n+1n+1 distinct points {c1,c2,...,cn+1}{c1,c2,...,cn+1} for which
p(cj)=0p(cj)=0 for all 1≤j≤n+1.1≤j≤n+1.
From the set of all such counterexamples, let p0p0 be one with minimum degree n0.n0.
That is, the claim in part (1) is true for any polynomial whose degree is smaller than n0.n0.
We write
p
0
(
z
)
=
∑
k
=
0
n
0
a
k
z
k
,
p
0
(
z
)
=
∑
k
=
0
n
0
a
k
z
k
,
(2)and we suppose that p0(cj)=0p0(cj)=0 for j=1j=1 to n0+1,n0+1,
where these ckck's are distinct complex numbers.
We use next the Root Theorem (part (d) of (Reference)) to write
p0(z)=(z-cn0+1)q(z),p0(z)=(z-cn0+1)q(z), where
q(z)=∑k=0n0-1bkzk.q(z)=∑k=0n0-1bkzk.
We have that qq
is a polynomial of degree n0-1n0-1 and the leading coefficient
an0an0 of p0p0 equals the leading coefficient bn0-1bn0-1 of q.q.
Note that for 1≤j≤n01≤j≤n0 we have
0
=
p
0
(
c
j
)
=
(
c
j
-
c
n
0
+
1
)
q
(
c
j
)
,
0
=
p
0
(
c
j
)
=
(
c
j
-
c
n
0
+
1
)
q
(
c
j
)
,
(3)which implies that q(cj)=0q(cj)=0 for 1≤j≤n0,1≤j≤n0,
since cj-cn0+1≠0.cj-cn0+1≠0.
But, since deg(q)<n0,deg(q)<n0, the nonzero polynomial qq
can not be a counterexample to part (1), implying that q(z)=0q(z)=0
for at most n0-1n0-1 distinct points.
We have arrived at a contradiction, and part (1) is proved.
Next, let rr be a polynomial for which r(z)=0r(z)=0 for an infinite number of
distinct points.
It follows from part (1) that rr cannot be a nonzero polynomial,
for in that case it would have a degree n≥0n≥0 and could be
0 for at most nn distinct points. Hence, rr is the
zero polynomial, and part (2) is proved.
Now, to see part (3), set r=p-q.r=p-q. Then rr is a polynomial for which r(z)=0r(z)=0
for infinitely many zz's. By part (2), it follows then that r(z)=0r(z)=0 for all z,z,
whence p(z)=q(z)p(z)=q(z) for all z.z.
Moreover, p-qp-q is the zero polynomial, all of whose
coefficients are 0, and this implies that the coefficients for
pp and qq are identical.
To prove the first inequality in part (4), suppose that |z|>1,|z|>1,
and from the backwards triangle inequality, note that
|
p
(
z
)
|
=
|
∑
k
=
0
n
c
k
z
k
|
=
|
z
|
n
|
∑
k
=
0
n
c
k
z
n
-
k
|
=
|
z
|
n
|
(
∑
k
=
0
n
-
1
c
k
z
n
-
k
)
+
c
n
|
≥
|
z
|
n
(
|
c
n
|
-
|
∑
k
=
0
n
-
1
c
k
z
n
-
k
|
)
≥
|
z
|
n
(
|
c
n
|
-
∑
k
=
0
n
-
1
|
c
k
|
|
z
|
n
-
k
)
≥
|
z
|
n
(
|
c
n
|
-
∑
k
=
0
n
-
1
|
c
k
|
|
z
|
)
≥
|
z
|
n
(
|
c
n
|
-
1
|
z
|
∑
k
=
0
n
-
1
|
c
k
|
)
.
|
p
(
z
)
|
=
|
∑
k
=
0
n
c
k
z
k
|
=
|
z
|
n
|
∑
k
=
0
n
c
k
z
n
-
k
|
=
|
z
|
n
|
(
∑
k
=
0
n
-
1
c
k
z
n
-
k
)
+
c
n
|
≥
|
z
|
n
(
|
c
n
|
-
|
∑
k
=
0
n
-
1
c
k
z
n
-
k
|
)
≥
|
z
|
n
(
|
c
n
|
-
∑
k
=
0
n
-
1
|
c
k
|
|
z
|
n
-
k
)
≥
|
z
|
n
(
|
c
n
|
-
∑
k
=
0
n
-
1
|
c
k
|
|
z
|
)
≥
|
z
|
n
(
|
c
n
|
-
1
|
z
|
∑
k
=
0
n
-
1
|
c
k
|
)
.
(4)Set BB equal to the constant (2/|cn|)∑j=0n-1|cj|.(2/|cn|)∑j=0n-1|cj|.
Then, replacing the 1/|z|1/|z| in the preceding calculation by 1/B,1/B, we obtain
|
p
(
z
)
|
≥
m
|
z
|
n
|
p
(
z
)
|
≥
m
|
z
|
n
(5)for every zz for which |z|≥B.|z|≥B.
This proves the first half of part (4).
To get the other half of part (4), suppose again that |z|>1.|z|>1.
We have
|
p
(
z
)
|
≤
∑
k
=
0
n
|
c
k
|
|
z
|
k
≤
∑
k
=
0
n
|
c
k
|
|
z
|
n
,
|
p
(
z
)
|
≤
∑
k
=
0
n
|
c
k
|
|
z
|
k
≤
∑
k
=
0
n
|
c
k
|
|
z
|
n
,
(6)so that we get the other half of part (4) by setting M=∑k=0n|ck|.M=∑k=0n|ck|.
Finally, to see part (5), suppose that there does exist a polynomial pp of degree nn
such that x=p(x)x=p(x) for all x≥0.x≥0.
Then x=(p(x))2x=(p(x))2 for all x≥0.x≥0.
Now p2p2 is a polynomial of degree 2n.2n.
By part (2), the two polynomials q(x)=xq(x)=x and (p(x))2(p(x))2 must be the same, implying that they have the same degree.
However, the degree of qq is 1, which is odd, and the
degree of p2p2 is 2n,2n, which is even.
Hence, we have arrived at a contradiction.
