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

2

+ 24T + 11 = 8(2T

2

+ 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

a2

≡ 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

a2

≡ 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: