2025/26 AHSMC Part 1 Problem 13

Write $n^2 + C = (n+m)^2$ with $m \ge 1$, so $C = 2mn + m^2 \le 50$, hence $n \le \dfrac{50-m^2}{2m}$. For each $m$ with $m^2 < 50$ (i.e. $m=1,\ldots,7$) the number of positive integers $n$ is $\bigl\lfloor(50-m^2)/(2m)\bigr\rfloor$, and $$\sum_{m=1}^{7} \Bigl\lfloor \frac{50-m^2}{2m} \Bigr\rfloor = 24+11+6+4+2+1+0 = 48.$$ The answer is (E).