Given that there is at least one solution to $a^{2} + 2b^{2} = p^{11}q^{13}$, find how many integers solutions there are.

Question

I cannot even begin this problem, given $ a, b \in \mathbb{Z}$ and $p,q$ odd prime numbers, given that there is a soltuion to the equation:

$a^{2} + 2b^{2} = p^{11}q^{13}$, find how many solutions there are.

I thought in see the left hand side as the norm of an element of $\mathbb{Z}[\sqrt-2]$, but I did not go too far. Does anyone have an idea to solve this type of question?

Thanks

Answer 1

Hint: Use unique factorization in $\mathbb Z[\sqrt{-2}]$. Show that $p$ and $q$ must factor in this domain.

Answer 2

Since $\mathbb{Z}[\sqrt{-2}]$ is a unique factorization domain, if the equation has solutions, then it has only one solution, ignoring multiplication by units. But since $\mathbb{Z}[\sqrt{-2}]$ has only two units ($1$ and $-1$), there can only be four solutions, which are different in a trivial way.

Furthermore, in order for the equation to have solutions, both $p$ and $q$ need to be primes that split in $\mathbb{Z}[\sqrt{-2}]$. For example, $p = 3 = (1 - \sqrt{-2})(1 + \sqrt{-2})$, $q = 11 = (3 - \sqrt{-2})(3 + \sqrt{-2})$. This is hopefully enough information for you to verify that $3^{11} 11^{13} = 2137242665^2 + 2 \times 879712646^2$.

Answer 3

This may be a bit late, but:

First note that $a=0$ or $b=0$ does not give any solutions, since $p,q$ odd, and that $R=\mathbb{Z}[\sqrt{2}]$ is a UFD with units $\{\pm1\}$. The number of solutions to $a^2+2b^2=p^{11}q^{13}$ is the cardinality of the set $\{\alpha\in\mathbb{Z}[\sqrt{-2}]\;|\;\;\alpha\tilde\alpha=p^{11}q^{13}\}$. Since $R$ is a quadratic field and a UFD, we have that a prime can be either inert, ramified or split, and can write $\alpha$ as a product of irreducibles, whose norm is either of $p,p^2,q,q^2$.

Claim: $p$ and $q$ are both split.

Proof: Since we're in a UFD, we can write $\alpha$ uniquely as a product of irreducibles, say $\alpha=\pm \pi_1^{r_1}...\pi_n^{r_n}$. Then $\alpha\tilde\alpha=\pi_1^{r_1}\tilde\pi_1^{r_1}...\pi_n^{r_n}\tilde\pi_n^{r_n}=p^{11}q^{13}$. If, say, $p$, is inert, and up to relabeling, $\pi_1=p$, then $\tilde\pi_1=p$ and we get that $p^{2r_1}=p^{11}$, which has no solution, contradiction. Similarly, if $p=\pi_1\tilde\pi_1$ is ramified, then $\pi_1\sim\tilde\pi_1\Rightarrow \pi_1=\pm\tilde\pi_1\Rightarrow p^{2r_1}=p^{11}$ again, no solution, so contradiction. This proves the claim, since there must exist at least one solution by the assumption.

Then write $p=\pi_p\tilde\pi_p$ and $q=\pi_q\tilde\pi_q$, so that $\alpha=\pm\pi_p^r\tilde\pi_p^s\pi_q^t\tilde\pi_q^u$ for positive integers $r,s,t,u$. Then $\alpha\tilde\alpha=p^{r+s}q^{t+u}=p^{11}q^{13}$, so we have that $s+r=11$ and $t+u=13$. So we have $2\cdot12\cdot14$ solutions by considering all possible values of $s,r,t,u$ that satisfy these equations and the $2$ comes from the two possible signs of $\alpha$.