Let h = the height of the right triangle (an integer).
Let c = the hypotenuse
Let l = the other leg
So l^2+h^2=c^2
I am trying to figure out, for instance, why there are 8 such triangles when h=12. I try to figure this out manually and I'm only finding (EDIT) 4:
5 12 13
9 12 15
16 12 20
35 12 37
Is there a general approach to finding these (other than factorizations of h^2)?
On
If you take (3,4,5) and multiply by 3 you get (9,12,15) but if you multiply with 4 you get (12,16,20). If your problem allows 0 and negative numbers or switching numbers then the count will change. For general solutions write $x^2+12^2=z^2$. This gives you $(z-x)*(z+x)=144$. factor 144 to two numbers, both have to be even (or both odd which does not happen here) to get integer answers, so you get $((z-x),(z+x))=(2,72),(4,36),(6,24),(8,18)$. First one gives $z=37,x=35$, next $z=20 x=16$, next $z=15,x=9$, next $z=13, x=5$. If you allow for exchanging legs you get 8. So the unfamiliar answer was $12^2+35^2=37^2$.
On
You might find this paper of interest: http://www.math.ou.edu/~dmccullough/teaching/pythagoras2.pdf
You can rewrite the Pythagorean theorem for your case as $144=c^2-\ell^2=(c-\ell)(c+\ell)$. There are lots of ways to factorize 144, which means lots of ways of writing it as $(c-\ell)(c+\ell)$.
Addendum: Just expanding a bit on some of what was covered in the comments. In response to Maesumi's answer, you make the remark,
If $h$ is odd, then, writing $h^2=xy$ with $x\lt h$ and $y\gt h$, the triples $(a,b,c)$ with $b=h$ are $((y-x)/2,h,(y+x)/2)$. If $h$ is even, then, writing $h^2=4wz$ with $w\lt h/2$ and $z\gt h/2$, the triples are $(z-w,h,z+w)$. This implies that a formula for the number of triangles of height $h$ is $$ \begin{cases} \frac{\sigma_0(h^2)-1}{2} & h\text{ odd,}\\ \frac{\sigma_0((h/2)^2)-1}{2} & h\text{ even,} \end{cases} $$ where $\sigma_0(n)$ is the number of divisors of $n$. Not that this helps you at all, since it's equivalent to what you're already doing. I gather that you are needing to compute this for a very large number of very large values of $h$. I'm not sure that there's any faster way to do this than by actually factorizing those values of $h$.
You've probably seen the Online Enyclopedia of Integer Sequences page related to your problem, http://oeis.org/A046079, but if not, it's worth having a look.
Addendum II: Filling in more details from the comments, which I muddled due to bad notation. You ask for a bound on the hypotenuse, $c$, given the height $h$. Let $\ell$ be the length of the other leg. The bound is different for $h$ even and for $h$ odd.
$h$ even: Since $c$ and $\ell$ must have the same parity if $h$ is even, we have $\ell\le c-2$ in this case, and so $$ h^2=c^2-\ell^2\ge c^2-(c-2)^2=4c-4. $$ Hence $c\le c_\text{max}=(h/2)^2+1$.
$h$ odd: Since $c$ and $\ell$ must have opposite parity when $h$ is odd, we have $\ell\le c-1$ in this case. From this it follows that $c\le c_\text{max}=(h^2+1)/2$.
Interestingly, there is always a triple with maximal hypotenuse since $c_\text{max}^2-h^2$ is always a perfect square.