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k ∈ B,weget n − k ∈ S. These are the two elements of S which
sum to n. V. The Waring Problem 51
j
Repeating Lemma 1 j times, then, leads to a summing of 2 copies
of S and a density of 1 − (1 − αð 2 j or more. Since this latter quantity,
for large enough j, will become bigger than 1 , Lemma 2 tells us
2
that 2 j+1 copies of S give us all the integers, just as Schnirelmann’s
theorem claims. Q.E.D.

C. Evaluation of Weyl Sums. Let b ∈ Z, b ø 0 and k ≤ N,
P (n) be a polynomial of degree k with real coefficients and leading
coefficient integral and prime to b, and let I be an interval of length
≤ N. Then

P (n) 1−k
e N 1+o(1) −2
b
b
n∈I
where the bound depends on k.

Here – as usual – we denote eàx) e 2πix .

We proceed by induction on k, which represents the degree of
P (n). It is clearly true for k 1, and generally we may write

P(n)
S e
b
n∈I
and may assume w.l.o.g. that I {1, 2, 3,...,N}. Thereby

N−1
P(n) − Pàn − jð
2
|S| e .
b
j −N+1 n∈{1,2,...,N}
n∈{j+1,j+2,...,j+N}
This inner sum involves a polynomial of degree (k − 1) but has a
leading coefficient which varies with j. If we count those j which
produce a denominator of d, which of course must divide b, then we
observe that this must appear roughly d times in an interval of length
b. So this number of j in the full interval of length 2N + 1 is roughly
(2N+1) d.
b