How many right angled triangles are there, all the sides of which are integers, having $2009^{12}$ as one of its shorter sides? Note that a triangle with sides a, b, c is the same as a triangle with sides b, a, c; where c is the hypotenuse.
I think I got the premise of the solution, but I dont know what I'm not counting
we need
$7^{24}41^{12} =(c-b)(c+b)$
There will always be a c+b for every c-b available, there are $\frac{25*13 + 1}{2} -1$ pairs of factors where$c+b > c-b$ = 163 possibilites. I am far from the answer but I don't know what I'm missing
By the formula for generating primitive Pythagorean triples $(a,b,c)$ we need to find all integers $m,n$ such that $$ a=2009^{12}=m^2-n^2,\; b=2mn,\; c=m^2+n^2. $$ So indeed, we need to count all possible different factorizations of $7^{24}41^{12}$ as $(m-n)(m+n)$.
Hint for your question: have a look at this sequence. So the number is $$ \frac{(2\cdot 24+1)(2\cdot 12+1)-1}{2}. $$