I was reading: https://homepages.warwick.ac.uk/~maslan/docs/T-seminar.pdf
Where they go over a proof that $x^2 - 5y^2 = 3z^2$ doesn't have integer solutions $(x,y,z)$ such that $z \ne 0$. Their proof has the following skeleton (outlined from the powerpoint but with more detail...)
$$ x^2 - 5y^2 \equiv 0 \mod 3$$ $$ x^2,y^2 \in \lbrace 0, 1 \rbrace \mod 3, \ 5 \equiv 2 \mod 3 \rightarrow x^2\equiv y^2 \equiv 0 \mod 3 $$ $$ \rightarrow x \equiv y \equiv 0 \mod 3 \rightarrow x^2 \equiv y^2 \equiv 0 \mod 9$$
$$ \rightarrow 3z^2 \equiv 0 \mod 9 \rightarrow z^2 \equiv 0 \mod 3 \rightarrow z^2 \equiv 0 \mod 9$$
$$ \rightarrow x^2 - 5y^2 \equiv 0 \mod 27$$
Now at this point it is implied that it is easy to "deduce" that this chain continues arbitrarily longer as infinite descent. But I reject that being obvious, in order for this "descent" to continue we have to prove that for each odd-power of 3 there doesn't exist a pair of quadratic residues $a,b$ other other than 0 such that $a=5b$. otherwise my step in line 2 won't be able to generalize.
So that leads to a natural conjecture: is it true that for all positive odd $n$ there doesn't exist non-zero quadratic residues $a,b \mod 3^n$ such that $a \equiv 5b$? How do I prove that? It seems like it requires induction but its not clear.
Of course that has a natural generalization too (since we could relabel our "classes" arbitrarily ):
$$ x^2 \equiv a \mod c \\ y^2 \equiv b \mod c \\ a = kb \\ \rightarrow \exists L | L^2 \equiv k \mod c$$