Romania — NMO Selection Tests for the Balkan and International Mathematical Olympiads

a.
We must show that there are infinitely many positive integer numbers $n$ such that the equation $y^2 = x^2 + (x+1)^2 + \dots + (x+n-1)^2$ has positive integral solutions. To this end, rewrite the equation in the form
$$y^2 = n\left(\left(x + \frac{n-1}{2}\right)^2 + \frac{(n^2-1)}{12}\right). \qquad (1)$$
We show that if $n$ is a square greater than $25$ not divisible by $2$ or $3$, then (1) has a positive but finite number of solutions in positive integers $x$ and $y$. Since $n$ is a perfect square, the expression in the outmost parentheses in (1) must be a perfect square $z^2$, i.e., there is a positive integer $z$ such that
$$z^2 - \left(x + \frac{n-1}{2}\right)^2 = \frac{(n^2-1)}{12}. \qquad (2)$$
Thus, $z \pm (x + (n-1)/2)$ must be complementary even divisors of $(n^2 - 1)/12$ differing by more than $n-1$; hence there are only finitely many solutions to (2) in positive integers. One such is obtained by taking
$$z - \left(x + \frac{n-1}{2}\right) = 2 \quad \text{and} \quad z + \left(x + \frac{n-1}{2}\right) = \frac{n^2-1}{24},$$
i.e., by letting $x$ and $z$ have the positive integer values
$$x = \frac{(n-25)(n+1)}{48} \quad \text{and} \quad z = 1 + \frac{(n^2-1)}{48}.$$

b.
Rewrite the equation $y^2 = x^2 + (x+1)^2 + \dots + (x+n-1)^2$ in the form
$$(2y)^2 - n(2x + n - 1)^2 = \frac{(n-1)n(n+1)}{3} \quad (3)$$
and let $T_n(x) = x^2 + (x+1)^2 + \dots + (x+n-1)^2$. We may assume $n > 1$. Since $T_n(x) = T_n(-x-n+1)$, we may further assume that $x_0 \ge -(n-1)/2$. Suppose $T_n(x_0) = y_0^2$, i.e., $(2y_0)^2 - n(2x_0 + n-1)^2 = (n-1)n(n+1)/3$, where we may assume $y_0 > 0$ (and hence $y_0 > (n-1)/2$). By the theory of the Pell equation, there are infinitely many pairs of positive integers $u, v$ such that
$$u^2 - nv^2 = 1. \quad (4)$$
Clearly, $u$ is odd if $n$ is even, so that $(n-1)(u-1)$ is even. We now use the identity
$$(2y_0 + (2x_0 + n - 1)\sqrt{n})(u + v\sqrt{n}) = 2y + (2x + n - 1)\sqrt{n}, \quad (5)$$
where
$$x = x_0u + y_0v + \frac{(n-1)(u-1)}{2} \quad \text{and} \quad y = y_0u + x_0nv + \frac{(n-1)nv}{2}. \quad (6)$$
Multiplying (5) by the identity obtained from (5) by replacing $\sqrt{n}$ by $-\sqrt{n}$, we find that if $x$ and $y$ are given by (6) and $u$ and $v$ satisfy (4), then
$$\begin{aligned} (2y)^2 - n(2x + n - 1)^2 &= ((2y_0)^2 - n(2x_0 + n - 1)^2)(u^2 - nv^2) \\ &= (2y_0)^2 - n(2x_0 + n - 1)^2 = \frac{(n - 1)n(n + 1)}{3}. \end{aligned}$$
Thus, if $x, y$ are given by (6), they satisfy (3), so that $T_n(x)$ is a perfect square. Further, since $x_0 \ge -(n-1)/2$ and $y_0 > (n-1)/2$, it follows that $y \ge y_0u > 0$ and $x \ge y_0v - u(n-1)/2 + (n-1)(u-1)/2 = y_0v - (n-1)/2 \ge y_0 - (n-1)/2 > 0$. This ends the proof.