Prove that $n$ is divisible by $6$.

Question

If the quadratic equations $x^2-mx+n=0$ and $x^2+mx-n=0$ both have integral roots, prove that $6|n$.

I've proved that $3|n$, and that $2|n$ for odd $m$, but I can't seem to prove it for even $m$.

Please help.

Answer 1

Hint: Since both eqaution have intergal solution we have $$m^2-4n=a^2\;\;\;{\rm and}\;\;\;\;m^2+4n =b^2$$ for some integers $a,b$.

So $2m^2 =a^2+b^2$ implies

a) if $3\mid m$ then $3\mid a$ and $3\mid b$ implies $3\mid a^2-b^2 =4n$ so $3\mid n$

b) if $3\nmid m$ then $a^2\equiv_3 1$ and $b^2\equiv_3 1$ implies $$4n =a^2-b^2 \equiv _3 0 \implies 3\mid 4n\implies 3\mid n$$

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Say $x_1$ and $x_2$ are solution to first equation. By Vieta formula $x_1+x_2 = m$ we see that if one is integer solution then so is the second. Now from $n=x_1x_2$ we get if $n$ is odd, then so are $x_1$ and $x_2$. Thus $m$ is even and so $a=2c$ and $b=2d$ from some integers $c,d$. Now we have $$2n =x^2-y^2=(x-y)(x+y)$$

which means that $x\equiv_2 y$. But then $2\mid n$.