Prime Numbers

9.8 Research problems 537
1.66), namely
Uk(a) =
N−1
x=0
e 2πiaxk /N .
Denote by rs(n) the number of representations of n as a sum of sk-th powers
in ZN. Prove that whereas
it also happens that
N−1
n=0
N−1
n=0
rs(n) =N s ,
rs(n) 2 = 1
N−1
N
|Uk(a)| 2s .
It is this last relation that allows some interesting bounds and conclusions.
In fact, the spectral sum of powers |U| 2s , if bounded above, will allow lower
bounds to be placed on the number of representable elements of ZN. In other
words, upper bounds on the spectral amplitude |U| effectively “control” the
representation counts across the ring, to analytic advantage.
Next, as an initial foray into the many research options, use the ideas and
results of Exercises 1.44, 1.66 to show that a positive constant c exists such
that for p prime, more than a fraction c of the elements of Zp aresumsoftwo
cubes. Admittedly, we have seen that the theory of elliptic curves completely
settles the two-cube question—even for rings ZN with N composite—in the
manner of Exercise 7.20, but the idea of the present exercise is to use the
convolution and spectral notions alone. How high can you force c for, say,
sufficiently large primes p? One way to proceed is first to show from the
“p 3/4 ” bound of Exercise 1.66 that every element of Zp isasumof5cubes,
then to obtain sharper results by employing the best-possible ”p 1/2 ” bound.
And what about this spectral approach for composite N? Inthiscaseonemay
employ, for appropriate Fourier indices a, an“N 2/3 ” bound (see for example
[Vaughan 1997, Theorem 4.2]).
Now try to find a simple proof of the theorem: If N is prime, then for
every k there exist positive constants ck, ɛk such that for a ≡ 0(modN) we
have
|Uk(a)|