The hypotenuse $c$ and one arm $a$ of a right triangle are consecutive integers. The square of the second arm is: 
$\textbf{(A)}\ ca\qquad \textbf{(B)}\ \frac{c}{a}\qquad \textbf{(C)}\ c+a\qquad \textbf{(D)}\ c-a\qquad  \textbf{(E)}\ \text{none of these}$
The sides of the triangle are $a,$ $\sqrt{c^2-a^2},$ and $c.$ We know that the hypotenuse and one leg are consecutive integers, so we can rewrite the side lengths as $a,$ $\sqrt{2a+1},$ and $a+1.$
Squaring the middle length gets $2a + 1 = a + c,$ so the answer is $\boxed{\textbf{(C)}}.$
-coolmath34