Proof related to quadratic equation

Question

Suppose that m and n are integers such that both the quadratic equations $x^2 + mx − n = 0$ and $x^2 − mx + n = 0$ have integer roots.

How to prove that n is divisible by 6?

Answer 1

As suggested in the comments, the quadratic formula tells you that $m^2-4n,m^2$ and $m^2+4n$ must be perfect squares.

Now, use that perfect squares are only $0,1 \mod 3$ to show that $3|n$.

Use that perfect squares are only $0,1,4,9 \mod 16$ to show that $2|n$.

EDIT:

Okay, as it seems not so easy to finish from here, I will do it for you:

First, we prove $3|n$.

Assume $3 \not \mid n$. Then $m^2,m^2-4n,m^2+4n$ all leave different residues $\mod 3$. So one of them must be $2 \mod 3$. Absurd! Hence $3|n$.

Now, we want to prove $2|n$.

Assume $2 \not \mid n$. Then $m^2,m^2-4n,m^2+4n$ all leave different residues $\mod 16$ but have the same residue $\mod 4$. But there are only two possible odd ($1$ and $9$) and two possible even ($0$ and $4$)residues of perfect squares $\mod 16$.

So again we derived a contradiction and conclude $2|n$ i.e. $6|n$. Hence the claim.