Prove that both $4m^2+17n^2$ and $4n^2+17m^2$ cannot be perfect squares for positive integers $m$ and $n$. Prove that both $4m^2+17n^2$ and $4n^2+17m^2$ cannot be perfect squares for positive integers $m$ and $n$.
I tried square bounding but didn't get very far.
Thanks for any help.
 A: The main thing about the sum of two squares is that, for any (positive) prime $$ q \equiv 3 \pmod 4,$$  if
$$ q | (x^2 + y^2)  $$ then actually $$ q|x,y $$ both. By induction, we get $$ q^{2k} \parallel (x^2 + y^2),  $$ where this notation means that $ q^{2k} | (x^2 + y^2)  $ but not $ q^{2k +1} | (x^2 + y^2)  .$ The notation is used only when the left hand side is a prime, otherwise it becomes ambiguous; it is a handy shorthand, though.  
In your problem statement, add the two together, you get
$$ 21 (m^2 + n^2) = x^2 + y^2.  $$ This is a contradiction, as 3 now divides the right hand side by an odd power. Same for 7.
Given any quadratic form $$ f(x,y) = a x^2 + b x y + c y^2, $$ abbreviated as $$ \langle a,b,c \rangle $$ the discriminant is $$   \Delta = b^2 - 4 a c.$$ For positive forms this is negative. For indefinite forms, this is positive. We discard the case when $\Delta$ is $0$ or a (positive) square, as those forms factor. 
For any (positive) odd prime $q$ with Legendre symbol $(\Delta | q) = -1,$ then, whenever $q$ divides $  a x^2 + b x y + c y^2, $ it follows that $$ q^{2k} \parallel (  a x^2 + b x y + c y^2).  $$ Worth proving yourself, in full detail. Caution: part of the hypothesis is that $\Delta \neq 0 \pmod q.$ That matters. Not sure how much more to say: there are thus two cases, $a \neq 0 \pmod q$ and $a \equiv 0 \pmod q,$ which is actually quite similar because _.
