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. Prove that $n$ is divisible by $6$.
Problem
Source: Indian RMO 2013 Paper 1 Problem 6
Tags: quadratics, modular arithmetic, number theory, Diophantine equation, number theory unsolved
the_boss
01.02.2014 19:41
official solution.......... http://olympiads.hbcse.tifr.res.in/uploads/crmo-2013-solutions-1
mavropnevma
01.02.2014 19:47
Look at the discriminants $\Delta_1 = m^2+4n = a^2$ and $\Delta_2 = m^2-4n = b^2$. Then $2m^2 = a^2+b^2$. The general solution of this Diophantine equation is $a= t(p^2+2pq-q^2)$, $b= t(p^2-2pq-q^2)$ ($m = t(p^2+q^2)$, with $\gcd(p,q)=1$). Therefore $8n = a^2-b^2 = 8t^2pq(p^2-q^2)$, so $n = t^2pq(p^2-q^2)$. However, $pq(p^2-q^2) \equiv 0 \pmod{6}$ is trivially verified.
aops29
03.12.2018 14:25
This is my solution:
Let \(a,b\) be the roots of \(x^2 + mx - n=0\) and let \(c,d\) be roots of \(x^2 - mx + n=0\). Vieta gives \[a^2 + b^2 + c^2 + d^2 = 2m^2\]If \(n\) is odd, then \(m\) must be even, and all of \(a,b,c,d\) must be odd. In that case, we have
\[a^2 + b^2 + c^2 + d^2 \equiv 0 \pmod{8}\]which is impossible, because \(a^2\equiv b^2 \equiv c^2 \equiv d^2 \equiv 1\pmod{8}\). So \(n\) is even. Now assume (for the sake of contradiction) that \(n\) is not divisible by \(3\). Then none of \(a,b,c,d\) is divisible by 3. Henceforth we have
\begin{align*}
a^2 + b^2 + c^2 + d^2 &\equiv 2m^2 \pmod{3} \\
4 &\equiv 2m^2 \pmod{3} \\
m^2 &\equiv 2 \pmod{3}
\end{align*}which is obviously impossible. Therefore, \(n\) is even and is divisible by \(3\), and the proof is finished.