TITLE: Proof that $x^4 - qy^4 = az^2$ has no integral solution
QUESTION [5 upvotes]: This is a question from Takashi Ono's book, Problem 1.45 to be exact.
The question is
Let $q$ be a prime such that $q = 1 \mod 8$ and $a$ be an integer such that $p^2\not\mid a$ for any prime $p$ and that $x^4 = a \mod q$
has no solution in $Z$. Prove that the equation $x^4 - qy^4 = az^2$ has no
integral solution other than x = y = z = 0.
I'm really stuck, this question comes after the section on quadratic reciprocity but not sure what I can do with it

REPLY [3 votes]: If there are solutions, then there are solutions with $x$, $y$, $z$ pairwise
coprime. If $p\mid x$, $y$, then $p^4\mid az^2$ and as $p^2\nmid a$ then
$p^2\mid z$. Then $(x/p,y/p,z/p^2)$ is also a solution. Similar
arguments work when $p\mid x,z$ and when $p\mid y,z$.
Suppose that $(x,y,z)$ form a 
 pairwise coprime solution to the equation.
Let $p$ be an odd prime dividing $z$. Then as $x^4\equiv qy^4\pmod p$
we get $\left(\frac qp\right)=1$ and then
$\left(\frac pq\right)=1$ by quadratic reciprocity.
As $\left(\frac {-1}q\right)=\left(\frac 2q\right)=1$ then
$\left(\frac zq\right)=1$.
There is $u$ with $z\equiv u^2\pmod q$. Then $x^4\equiv au^4\pmod q$,
and as both sides are nonzero modulo $q$, then $a$ is a quartic residue,
a contradiction.
The example $x^4-17y^2=2z^2$ is a counterexample to the Hasse principle
to quartics and is due to Reichardt.