9. Primes represented by general polynomials 27
namely 11. We replace T by 4T + 3 and so obtain from F the polynomial
G(T) = F (4T + 3) = 16T
+ 24T + 11 = 8(2T
+ 3T) + 11.
Then every prime divisor of G belongs to either the residue class 1 mod 8 or
3 mod 8. Moreover, for each positive integer n, there is at least one prime
p ≡ 3 (mod 8) for which p | G(n), since G(n) ≡ 3 (mod 8). We will show
that G (and hence also F ) must have infinitely many prime divisors from
the residue class 3 mod 8. Suppose otherwise, and let p1,p2,...,pk be a
complete list of the prime divisors p ≡ 3 (mod 8) of G. For each pi, choose
an integer ni for which G(ni) ≡ 0 (mod pi). (This is possible since G has at
most two roots modulo pi.) If n is a positive integer chosen by the Chinese
remainder theorem to satisfy n ≡ ni (mod pi) for all 1 ≤ i ≤ k, then G(n)
cannot be divisible by any of p1,...,pk. So G(n) must have a prime divisor
from the residue class 3 mod 8 other than p1,...,pk, a contradiction.
Example. Since every integer a coprime to 24 satisfies
≡ 1 (mod 24), it
is in principle possible to give an algebraic proof of Dirichlet’s theorem for
progressions with common difference 24. The details in this case have been
completely worked out by Bateman & Low [BL65]. We leave to the reader
the task of showing that 24 is the largest modulus m with the property that
≡ 1 (mod m) for each a coprime to m.
9.2. Hypothesis H.
I do not mean to deny that there are mathematical truths,
morally certain, which defy and will probably to the end of
time continue to defy proof, as, e.g., that every indecom-
posable polynomial function must represent an infinitude of
primes. – J. J. Sylvester [Syl88]
There are two natural directions we might head in if we hope to gen-
eralize Dirichlet’s result: First, we might inquire about simultaneous prime
values of several linear polynomials. One has to be careful here, of course.
For example, we cannot hope that there are infinitely many n for which
both n and n + 1 are prime, because one of these two numbers is always
even! However, if instead of n and n + 1 we consider n and n + 2, then this
obstruction disappears, and we arrive at the following famous conjecture:
Conjecture 1.23 (Twin prime conjecture). There are infinitely many nat-
ural numbers n for which both n and n + 2 are prime.
Alternatively, we might accept the restriction of working with a single
polynomial, but hope to treat polynomials of higher degree. The following
conjecture of Euler, which appears in correspondence with Goldbach, fits
nicely into this framework: