Prime Numbers

1.5 Exercises 71
(2) The infinitude of prime triples in arithmetic progression (see Exercises
1.41, 1.42) is equivalent to the divergence as N →∞of
1
0
E 2 N(t)EN(−2t) dt.
(3) The (binary) Goldbach conjecture is equivalent to
1
e
0
−2πitN E 2 N(t) dt = 0
for even N>2, and the ternary Goldbach conjecture is equivalent to
for odd N>5.
1
e
0
−2πitN E 3 N(t) dt = 0
(4) The infinitude of Sophie Germain primes (i.e., primes p such that 2p +1
is likewise prime) is equivalent to the divergence as N →∞of
1
0
e 2πit EN(2t)EN(−t) dt.
1.68. We mentioned in Section 1.4.4 that there is a connection between
exponential sums and the singular series Θ arising in the Vinogradov
resolution (1.12) for the ternary Goldbach problem. Prove that the Euler
product form for Θ(n) converges (what about the case n even?), and is equal
to an absolutely convergent sum, namely,
Θ(n) =
∞
q=1
µ(q)
ϕ 3 (q) cq(n),
where the Ramanujan sum cq is defined in (1.37). It is helpful to observe
that µ, ϕ, c are all multiplicative, the latter function in the sense that if
gcd(a, b) = 1, then ca(n)cb(n) =cab(n). Show also that for sufficiently large
B in the assignment Q =ln B n, the sum (1.40) being only to Q (and not to
∞) causes negligible error in the overall ternary Goldbach estimate.
Next, derive the representation count, call it Rs(n), for n the sum of s
primes, in the following way. It is known that for s>2, n ≡ s (mod 2),
Rs(n) = Θs(n) n
(s − 1)!
s−1
ln s n
1+O
ln ln n
ln n
,
where now the general singular series is given from exponential-sum theory as
Θs(n) =
∞
q=1
µ s (q)
ϕ s (q) cq(n).