Past exam of the mathematics course of the University of Cambridge / 2017 / ii / Paper 4 / 10G (Number Theory)

Dirichlet theorem on primes in arithmetic progressions states that if positive integer $q$ and integer $a$ satisfy $\gcd (a,q)=1$, then infinitely many prime numbers satisfyThe coprimality hypothesis is essential: if $a$ and $q$ have a common factor greater than one, a prime in the progression must divide that factor, giving at most finitely many possibilities.

Write an integral binary quadratic form as $Q(u,v)=A{u}^2+Buv+C{v}^2$, with discriminant of a binary quadratic form $d=B^2-4AC$. If $Q(u,v)=p$ is prime, then $\gcd (u,v)=1$: any common divisor would have its square dividing $p$. Complete $(u,v)$ to a matrix in ${\mathrm{SL}}_2(\mathbb{Z})$ using Bezout identity. After the associated integral change of variables, the properly equivalent form has leading coefficient $p$, say $[p,b,c]$. Its discriminant of a binary quadratic form gives $d=b^2-4pc$, hence $b^2\equiv d(\mathrm{mod}p)$.

Conversely suppose $b_0^2\equiv d(\mathrm{mod}p)$. Since $d$ is an integral-form discriminant of a binary quadratic form, $d\equiv 0$ or $1(\mathrm{mod}4)$. Choose $b\equiv b_0(\mathrm{mod}p)$ with even integer parity for $d\equiv 0(\mathrm{mod}4)$ and odd integer parity for $d\equiv 1(\mathrm{mod}4)$. This is possible because $p$ is odd. Then $b^2-d$ is divisible by $4p$, andis an integral form of discriminant of a binary quadratic form $d$ representing $p=Q(1,0)$. This proves the discriminant criterion for prime representation by a binary quadratic form, including the case $p\mid d$; no assumption that $d$ is a fundamental discriminant was used.

Both forms have discriminant of a binary quadratic form $-60$. The reduction algorithm for a positive definite binary quadratic form gives $|b|\le a\le c$ and $a\le \sqrt{60/3}<5$ for a reduced positive definite representative. Enumerating $a=1,2,3,4$ givesThe second and fourth forms are imprimitive and represent only even integers. Every odd prime represented by a form of discriminant of a binary quadratic form $-60$ must therefore be represented by $f$ or $g$.

For a prime $p\nmid 30$, quadratic reciprocity givesEvery prime $p\equiv 1(\mathrm{mod}15)$ has this symbol equal to one and hence is represented by one of the two primitive forms, by part (b). But an odd prime represented by $g$ and different from three is $2(\mathrm{mod}3)$, so these primes must be represented by $f$. Similarly, primes $p\equiv 2(\mathrm{mod}15)$ have both symbols equal to $-1$, and are represented by $g$, since a prime represented by $f$ is $1(\mathrm{mod}3)$. Both residue classes are coprime to 15, so Dirichlet theorem on primes in arithmetic progressions proves each form represents infinitely many primes.

The same modulo-three distinction excludes every prime other than three from being represented by both. The form $f$ does not represent three; $g$ represents three and five but neither is represented by $f$, and two is represented by neither. Therefore there are no primes represented by both forms.