Prove that $p^2 - 4qr$ ($p,q,r$ odd natural numbers) is never a perfect square The givens for the question: $p, q, r$ are odd natural numbers.
We need to prove that $p^2 - 4qr$ is never a perfect square.
Inspecting a few examples it seems to be true, but I have no idea where to start. What approach would one use?
 A: This is equivalent to proving that for odd $m$, if $4|n^2-m^2$ then $8|n^2-m^2$.
This is easy if we notice $n^2-m^2=(n+m)(n-m)$, and if one of them is even, both are even, and also, one is a multiple of $4$.
A: If $p,q,r$ are odd then $p^2-4qr$ is odd.  So if $p^2-4qr=s^2$ then $s$ must be odd.  But $4qr=(p-s)(p+s)$  The LHS is divisible by $4$ and not $8$, the RHS is divisible by $8$ because one of $p+s$ or $p-s$ is divisible by $2$ and the other by $4$.
A: A little culture. If $p^2 - 4 q r$ were a perfect square, then we would be able to factor
$$  qx^2 + p x + r  $$ over the integers as
$$  qx^2 + p x + r = ( q_1x + r_1) (q_2 x + r_2); $$
see Prove that if $b^2-4ac=k^2$ then $ax^2+bx+c$ is factorizable or
Formula for factorization of a Quadratic Equation?
However,
$$  ( q_1x + r_1) (q_2 x + r_2) = q_1 q_2 x^2 + (q_1 r_2 + r_1 q_2)x + r_1 r_2,  $$
while $(q_1 r_2 + r_1 q_2)$ is even, so it cannot be equal to $p.$
A: And now a strange solution. Assuming that $p^2-4qr$ is a square, the polynomial
$$s(x)= qx^2 + px + r $$
has to be reducible over $\mathbb{Q}$. However, $x^2+x+1$ is irreducible over $\mathbb{F}_2$, hence that cannot happen.
A: If it's square, $ x^2\! + p x\! +\! qr\,$ has odd integer roots (factors of odd $qr),\,$ contra root sum $ {-}p\,$ is odd
A: Suppose to the contrary that $p^2-4qr=x^2$, where $x$ is an integer. Then $p^2-x^2=4qr$.
It is clear that $x$ must be odd. So $p^2\equiv 1\pmod{8}$ and $x^2\equiv 1\pmod{8}$, and therefore $p^2-x^2$ is divisible by $8$. But $4qr$ is not divisible by $8$.
A: $p^2$ is congruent to $1$ mod $8$ (it is equal to $8k+1$ for some $k$) but
$4qr$ is congruent to $4$ mod $8$ (it is $8m+4$ for some $m$).  
Therefore the difference of those two expressions is $(1 - 4) = 5$ mod $8$ which is not the value of any perfect square mod $8$.
