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C
C
C
2
2
2
2 , and finally
2 ,A (r ) ≥ Pàr ) +
2 ,A (r ) ≥ð M +
1−r
1−r
1−r
2 2 Erd˝ os–Fuchs Theorem 35
C
2
A(r ) ≥ −M + , à 10)
1 − r 2
a rate of growth which flatly contradicts (9) and so gives our desired
contradiction.
If this proof seems like just so much sleight of hand, let us ob-
serve what is “really” going on. We find ourselves with a set A
2 C
whose r i (n) is “almost” constant and this means that A (z) ≈ .
1−z
On the one hand, this forces A(z) to be large on the positive axis

2 C
A(r )> √ , and, on the other hand Parseval says that the
1−r 2

2 2 C (being fairly small except near
1−z
integral of |A (z)| is A(r ) and
2
1) has a small integral, only O(log 1 ). (So A(r )<C log 1 ).

1−r 1−r
2
In cruder terms, Parseval tells us that A (z) is large on average,
so it must be large elsewhere than just near z 1, and so it cannot
really be like C . (Note that the “elsewhere” in the earlier r + (n)
1−z
problem was the locale of −1, and so even that argument seems to
be in this spirit.)
So let us turn to the Erd˝ os–Fuchs theorem with the same strategy
2 C
in mind, viz., to bound A(r ) below by √ for obvious reasons
1−r 2
and then to bound it above by Parseval considerations.
Erd˝ os–Fuchs Theorem
We assume the A is a set for which
α
r(0) + r(1) + ··· + r(n) Càn + 1) + O(n ), C > 0,à 11)
1
and we wish to deduce that α ≥ . As usual, we introduce the
4
a 2 n
generating function A(z) z , so that A (z) r(n)z ,
a∈A

n
2
and therefore 1 A (z) [r(0) + r(1) + ··· + r(n)]z . Since
1−z
n 1
(n + 1)z 2 our hypothesis (11) can be written as
(1−zð
∞
1 2 C n α
A (z) + a n z , a n O(n ),
1 − z (1 − zð 2
n 0