Twentieth IMAR Mathematical Competition

Suppose, if possible, that $x+y$ is divisible by $2n+1$. Note that $2n+1$ and $2n^2-1$ are relatively prime, as $(2n-1)(2n+1)-2(2n^2-1)=1$. Hence $2n+1$ is coprime to $\gcd (x,y)$, so $x/\gcd (x,y)$ and $y/\gcd (x,y)$ are coprime divisors of $2n^2-1$ whose sum is divisible by $2n+1$.

To reach a contradiction, we may and will therefore assume $x$ coprime to $y$. Then $2n^2-1$ is divisible by $xy$, say, $2n^2-1=kxy$; write $x+y=\ell (2n+1)$. Express $n$ from this latter and plug it into the former to get $kxy=\frac{1}{2}{\left(\frac{x+y}{\ell }-1\right)}^2-1$; alternatively, but equivalently, $2k{\ell }^{2}xy=(x+y-\ell {)}^2-2{\ell }^2$.

Fix $k$ and $\ell$ to regard the above equality as an equation in positive integers $x$ and $y$. By assumption, it has at least one solution. Consider one such with a minimal $x+y$. Without loss of generality, assume $x\ge y$.

Rewrite the equation as a quadratic in $x$, $x^2+2(y-\ell -k{\ell }^{2}y)x+(y-\ell {)}^2-2{\ell }^2=0$, and consider the other root $x^{\prime }$. Then $x+x^{\prime }=2(k{\ell }^{2}y-y+\ell )$ and $x{x}^{\prime }=(y-\ell {)}^2-2{\ell }^2$. The former shows that $x^{\prime }$ is also integer. The latter implies $x^{\prime }<0$: Clearly, $x^{\prime }\ne 0$, as $(y-\ell {)}^2=2{\ell }^2$ cannot hold in integers; and if $x^{\prime }>0$, then $x^{\prime }y\le x{x}^{\prime }=(y-\ell {)}^2-2{\ell }^2=y^2-\ell (2y+\ell )<y^2$, so $x^{\prime }<y$, whence $x^{\prime }+y<2y\le x+y$, contradicting the minimality of $x+y$.

Thus, $(y-\ell {)}^2<2{\ell }^2$ and $x$ divides $2{\ell }^2-(y-\ell {)}^2$, so $x\le 2{\ell }^2$. On the other hand, as $x^{\prime }$ is negative, $x>x+x^{\prime }=2((k{\ell }^2-1)y+\ell )\ge 2({\ell }^2+\ell -1)\ge 2{\ell }^2$. This is a contradiction, so the sum $x+y$ is not divisible by $2n+1$, as required.