Problem 1946

We will prove that $p = 23$ is the unique solution. Subtracting the second equality from the first one, we get $p(p-1) = 2(y-x)(y+x)$. From the first equality $p$ is odd, thus $p \ne 2$ and $p \mid (y-x)(y+x)$. Clearly $y > x$, so assuming $p \mid (y-x)$, we get $p \le y-x < y+x$. Therefore,
$$2(y-x)(y+x) > 2p^2 > p^2 > p(p-1) \text{ - a contradiction.} \quad \implies \quad p \mid (y+x).$$

Assuming $y \ge p$, from the second equation it follows $49 = 2y^2 - p^2 \ge p^2$, i.e., $p \le 7$. Since $26 = \frac{3+49}{2}$, $27 = \frac{5+49}{2}$, $28 = \frac{7+49}{2}$ are not perfect squares, none of those values is a solution. For the remaining case $y < p$, we have $x+y < 2y < 2p$, thus $x+y = p$. So $y-x = p-2x$ and
$$p-1 = 2(y-x) = 2(p-2x) \Leftrightarrow p = 4x-1 \Leftrightarrow 2x^2 - 4x - 48 = 0 \Leftrightarrow x_{1,2} = 1 \pm 5.$$

Finally, since $-4 < 0$, the only solution is $x = 6$, $p = 4 \cdot 6 - 1 = 23$ and $y = p - x = 17$. Direct check confirms that this is indeed a solution.