$32 \cdot 62$ A right-angled triangle has one leg of length a prime number $n$, and the other two sides are certain natural numbers. Then its perimeter is
(A) $n^{2}+1$.
(B) $n^{2}-1$.
(C) $n^{2}+n$.
(D) None of the above answers is correct.
(2nd "Jinyun Cup" Junior High School Mathematics Invitational, 1985)
[Solution] Let the other leg and the hypotenuse of the right triangle be $x$ and $y$, respectively. By the Pythagorean theorem, we have
$y^{2}-x^{2}=n^{2}$, which can be rewritten as $(y-x)(y+x)=n^{2}$.
Solving this, we get $x=\frac{n^{2}-1}{2}, y=\frac{n^{2}+1}{2}$.

The perimeter of the triangle is $n+\frac{n^{2}-1}{2}+\frac{n^{2}+1}{2}=n+n^{2}$.
Therefore, the answer is (C).