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I. The Idea of Analytic Number Theory
8
Finally for r + (n), we must add A(z ) to this result to reinstate the

case of a a , and we obtain 2
1
n 2 2
r + (n)z [A (z) + A(z )]. à 9)
2
Can r(n) be “constant?”
Is it possible to design a nontrivial set A, so that, say, r + (n) is the same
for all n? The answer is NO, for we would have to have 0 ∈ A. And
then 1 ∈ A, else r + (1) ø r + (0). And then 2 /∈ A, else r + (2) 2.
And then 3 ∈ A, else r + (3) 0 (whereas r + (1) 1), then 4 /∈ A,
else r + (4) 2. Continuing in this manner, we find 5 ∈ A. But now
we are stymied since now 6 1 + 5, 6 3 + 3, and r + (6) 2.
The suspicion arises, though, that this impossibility may just be
a quirk of “small” numbers. Couldn’t A be designed so that, except
for some misbehavior at the beginning, r + (n) constant?
We will analyze this question by using generating functions. So,
using (9), the question reduces to whether there is an infinite set A
for which
1 2 2 C
[A (z) + A(z )] Pàz) + , à 10)
2 1 − z
Pàz) is a polynomial.
+
Answer: No. Just look what happens if we let z → (−1) . Clearly
2
Pàz) and C remain bounded, A (z) remains nonnegative, and
1−z
2
A(z ) goes to Aà 1) ∞, a contradiction.

A Splitting Problem

Can we split the nonnegative integers in two sets A and B so that
every integer n is expressible in the same number of ways as the
sum of two distinct members of A, as it is as the sum of two distinct
members of B?
If we experiment a bit, before we get down to business, and begin
by placing 0 ∈ A, then 1 ∈ B, else 1 would be expressible as