How to prove this modular problem?

Question

Prove that if $n^2+m$ and $n^2-m$ are perfect squares, them $m$ is divisible by $24.$ How to solving this problem?

Answer 1

Let $n^2+m = p^2, n^2-m = q^2$. If $n \equiv 0 \pmod 2$, then looking at quadraitc residues modulo $4$, we need $m \equiv 0, 1 \pmod 4$ from the first equation and $m \equiv 0, 3 \pmod 4$ from the second equation. Similarly, if $n$ is odd, then $m \equiv 0, 3 \pmod 4$ from the first equation and $m \equiv 0, 1 \pmod 4$ form the second equation.

Therefore, $n,p$ and $n,q$ have the same parity. Since $(n-p)(n+p) = m, $ we have $8|m$ since we have $2|n-p, n+p$ and $4$ divides only one of $n-p, n+p$.

Then looking at the equations $\pmod 3$, we have that if $n^2 \equiv 0 \pmod 3$, then $m \equiv 0$ from looking at both equations and if $n^2 \equiv 1 \pmod 3$, then $m \equiv 0 $ again. Since we have covered all residues modulo $3$, we can say $3|m$ and combining all of our results, $24|m$ as desired.

Answer 2

Hint $\ j^2+m = n^2,\ n^2+m = k^2\,\Rightarrow\, (k+j)^2+(k-j)^2 = (2n)^2$ is a Pythagorean Triple (PT) whose triangle has area $(k^2-j^2)/2 = m.\,$ Every primitive PT has area divisible by $\,\color{#c00}6\,$ and odd hypotenuse. Ours has even hypotenuse $\,2n,\,$ so it is an even scaling of a primitive triple, therefore our area gains a factor of $\,\color{#0a0}2\,$ from each leg, hence its area $\,m\,$ is a multiple of $\ \color{#0a0}{2\cdot 2}\cdot\color{#c00}6 =24$.