Question: A right triangle has all three sides integer lengths. One side has length 12. What are the possibilities for the lengths of the other two sides? Give a proof to show that you have found all possibilities.

EDIT: I figured out that there are a total of 4 combinations for a side with length 12.

$$a^2= c^2 -b^2$$

factoring the right side gives

$$a^2 = (c+b)(c-b)$$

so from there I just looked for values of c and b which would make the equation true. And I got: $(37,35), (13,5), (15,9), (20,16)$.

My only problem now is proving that these are all the possibilities. I have an intuition as to why that's true but I don't know how to go about a full proof. I'm guessing I need to make use of the fact that all sides are to be integer lengths and that $12$ itself cannot be equal to $c$.

I know that if I were to try values for values for $b$ and $c$ whose absolute value difference is greater than 8, then the equation would not hold true.

Ex:

$(37,35)$ has a difference of $2$ and works.

$(13,5)$ has a difference of $8$ and works.

$(15,9)$ has a difference of $6$ and works.

$(20,16)$ has a difference of $4$ and woks.

But if I were to pick any pair with an absolute difference greater than 8 it would not work. Can I just prove this is true with a couple of examples? Or do I need a full generic proof ? If so, how would I go about it?

Answer: 
If $$144=(c+b)(c-b)$$ we know that $c+b$ is a divisor of $144$. So there are only finitely many pairs of $(c,b)$ we have to investigate. $(c+b,c-b)$ are from the set $$\{(1,144),(2,72),(3,48),(4,36),(6,24),\ldots,(144,1)\}$$. But we know that $c+b \gt c-b$ and that both $c+b$ and $c-b$ are even. Because $12\cdot 12=144$ we have $c-b<12$ and $c+b>12$. The only even divisors of $144$ are $\{2,4,6,8\}$. So your solutions are the only ones.

But the side of length $12$ can be the hypotenuse $c$. So you should check the equation $$12^2=a^2+b^2$$ too. Again only finitely many pairs are to check. You can assume that $a \ge b $ and therefore $a \ge \sqrt{\frac{144}{2}} = 8.\cdots$. So you have only to check $a \in \{9,10,11\}$.
[Wolfram MathWorld](http://mathworld.wolfram.com/PythagoreanTriple.html) gives the number of ways in which a number $n$ can be a leg (other than the hypotenuse) of a primitive or non-primitive right triangle. For $n=12$, the number is $4$. It also gives the number of ways in which a number $n$ can be the hypotenuse of a primitive or non-primitive right triangle. For $n=12$, the number is $0$. So the four triples you found are the only ones.