Show that $a^2 - 2021b^2 = 13$ has no integer solutions

Question

Show that $a^2 - 2021b^2 = 13$ has no integer solutions.

I'm no number theory expert, but here's what I've thought of so far. If there exist $a,b$ that satisfy the above equation, then there must also exist $a,b$ that satisfy $$a^2 \bmod 4 = b^2\bmod 4 + 1$$ A square is $0$ or $1 \bmod 4$, so we get $a^2\equiv 1\bmod 4$ and $b^2 \equiv 0 \bmod 4$. Thus, $2|b$ and $a^2 = 4k+1$. I do not see a contradiction yet. What else should I try?

Also tried to use the fact that squares have units digit only $0,1,4,5,6,9$, so if $a,b$ exist that satisfy the equation, we must have $a^2$ ending in $\{0,1,4,5,6,9\} \cap \{3,4,7,8,9,2\} = \{4,9\}$. Moreover, we must have $b^2$ ending in $\{0,1,4,5,6,9\} \cap \{7, 8, 1,2,3,6\} = \{1,6\}$.

If $b^2$ ends in $1$, then $a^2$ ends in $4$ - but we just saw that $a^2 = 4k+1$, so this is not a possibility.

If $b^2$ ends in $6$, then $a^2$ ends in $9$. How do I contradict this? Thank you!

Answer 1

$a^2=13+2021b^2 \Rightarrow a^2 \equiv 6b^2 \mod 13 \iff \Big(\frac{a}{b}\Big)^2 \equiv 6 \mod 13 $ for $b\neq0$. But as $\genfrac(){}{}{6}{13}=-1$ then $6$ is not a quadratic residue so the equation does not have solutions unless $a \equiv b \mod 13$ and $a \equiv 0 \mod 13$. But even in this case, $13^2$ divides the LHS but does not divide the RHS.

Answer 2

A slightly different alternative that does not require separate consideration of zero residues:

Because any odd number $n$ is a product of only odd primes, quadratic reciprocity implies $(p|n)=(n|p)$ for any $4m+1$ prime $p$. Thus $(13|2021)=(2021|13)=(6|13)=-1$, and then we can't have $a^2\equiv13\bmod 2021$.