Problem 1586

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, $2k\ell^2xy = (x + y - \ell)^2 - 2\ell^2$.

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 $x$, $x^2+2(y-\ell-k\ell^2y)x+(y-\ell)^2-2\ell^2=0$, and consider the other root $x'$. Then $x+x'=2(k\ell^2y-y+\ell)$ and $xx' = (y-\ell)^2-2\ell^2$. The former shows that $x'$ is also integer. The latter implies $x' < 0$: Clearly, $x' \ne 0$, as $(y-\ell)^2 = 2\ell^2$ cannot hold in integers; and if $x' > 0$, then $x'y \le xx' = (y-\ell)^2 - 2\ell^2 = y^2 - \ell(2y+\ell) < y^2$, so $x' < y$, whence $x' + y < 2y \le x + y$, 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 $x'$ is negative, $x > x + x' = 2((k\ell^2 - 1)y + \ell) \ge 2(\ell^2 + \ell - 1) \ge 2\ell^2$. This is a contradiction, so the sum $x + y$ is not divisible by $2n + 1$, as required.