Find the number of non-congruent right triangles, where all the sides are positive integers, and one of the legs is $25$.

1 answer

Let $x$ denote the hypotenuse of the right triangle, and let $y$ denote the other leg. Because of the Pythagorean Theorem, $x^2 = 625 + y^2$. Completing the square, we get $x^2 - y^2 = (x + y)(x - y) = 625$. We note that $625 = 25^2$. We also note that because $x$ and $y$ are sides of the triangle, $x + y > x - y$. This means that $(x, y)$ and $(y, x)$ give us the same triangle, so we only need to consider one case.

We now proceed with casework on the prime factors of $625 = 5^4$.

Case 1: $x - y = 1$ and $x + y = 625$. This implies $x = 313, y = 312$. We can confirm these are in fact positive integers that satisfy the original equation. Thus, there is $1$ triangle in this case.

Case 2: $x - y = 5$ and $x + y = 125$. This implies $x = 65, y = 60$. Again, we confirm these are in fact positive integers that satisfy the original equation. Therefore there is $1$ triangle in this case as well.

Case 3: $x - y = 25$ and $x + y = 25$. The two equations give us $x = 25, y = 0$. However, one of the legs cannot be $0$. Therefore there are $\boxed{2}$ triangles overall.