Find the number of non-congruent right triangles, where all the sides are positive integers, and one of the legs is $162$.
1 answer
The side-lengths of a right triangle are given by $\{162, a, b\}$, where $a$ and $b$ are integers satisfying the Pythagorean Theorem $162^2 + a^2 = b^2$. Then, $b^2 - a^2 = (b+a)(b-a) = (b-a) \cdot 324 = 162^2 = 2^4 \cdot 3^4 \cdot 9^2$. Thus, $b-a$, $b+a$, and $b$ are a set of three factors of $2^4 \cdot 3^4 \cdot 9^2$. Realize that any factor $x$ of $2^4 \cdot 3^4 \cdot 9^2$ can be uniquely expressed such that it divides $2^4 \cdot 3^2 \cdot 9^2$, $b-a = x$, and $b+a = \frac{324^2}{x}$. Thus, there are $6 \cdot 3 \cdot 3 = 54$ possibilities. However, since $b-a \ne b+a$, and one must be subtracted twice, the answer is $\boxed{53}\ \mathbf{(C)}$.