HMMT February, 2018

Solution:
Let $n = p \cdot q$ for primes $p < q$. If $n$ is the second largest side of a right triangle there exist integers $c, a$ such that $a < p q$ and $(p q)^2 = c^2 - a^2 = (c-a)(c+a)$. Since $c-a < c+a$ there are three cases for the values of $c-a, c+a$, and in each case we determine when $a < p q$.

a. $c-a = 1$ and $c+a = p^2 q^2$ : Then $a = \frac{p^2 q^2 - 1}{2} > p q$, so there are no solutions.

b. $c-a = p$ and $c+a = p q^2$ : Then $a = \frac{p q^2 - p}{2} > p q$.

c. $c-a = p^2$ and $c+a = q^2$. Then $a = \frac{q^2 - p^2}{2}$ which we require to be less than $p q$. This is equivalent to
$$\begin{aligned} \frac{q^2 - p^2}{2} & < p q \\ q^2 & < 2 p q + p^2 \\ 2 q^2 & < (q + p)^2 \\ \sqrt{2} q & < q + p \\ (\sqrt{2} - 1) q & < p < q \end{aligned}$$
So the problem is equivalent to finding the number of distinct prime pairs $(p, q)$ such that $p q < 300$ and $(\sqrt{2} - 1) q < p < q$. There are 13 such pairs:
$$\{(3,5), (3,7), (5,7), (5,11), (7,11), (7,13), (11,13), (11,17), (11,19), (11,23), (13,17), (13,19), (13,23)\}$$
and $13 \cdot 23 = 299$ which is the biggest such pair.
The most interesting borderline case are $(3,7): \frac{3}{7} \approx .42 > \sqrt{2} - 1$, which leads to the $(20,21,29)$ triangle, $(5,13): \frac{5}{13} \approx .385 < \sqrt{2} - 1$, which leads to the $(65,72,97)$ triangle, and $(7,17): \frac{7}{17} \approx .411 < \sqrt{2} - 1$ which leads to the $(119,120,169)$ right triangle.