TITLE: If positive integers $a$, $b$, $c$ satisfy $\frac1{a^2}+\frac1{b^2}=\frac1{c^2}$, then the sum of all values of $a\leq 100$ is ...
QUESTION [7 upvotes]: I'm struggling to solve the following problem. I would like hint (just a hint, not a full solution please) on how to solve it: 

The positive integers $a$, $b$, and $c$ satisfy $$\dfrac1{a^2}+\dfrac1{b^2}=\dfrac1{c^2}$$
  The sunm of all possible $a\leq 100$ is ...
A) $315\quad$ B) $615\quad$ C) $680\quad$ D) $550\quad$ E) $620$
(Source: 2005 Cayley (Grade 10), #25) 
  Primary Topics: Number Sense 
  Secondary Topics: Counting | Fractions/Ratios

(original problem image) 
What I've done so far is that I've rearranged $1/a^2 + 1/b^2 = 1/c^2$ to get $a^2 + b^2 = (ab/c)^2$. Then this means that $a$, $b$ and $ab/c$ are pythagorean triples, because 
$$(integer)^2 + (integer)^2 = (integer)^2$$ 
But I'm not sure how to proceed from there, I'd really appreciate a hint.
Thanks in advance!

REPLY [3 votes]: Let $(a,b)=d$
WLOG $\dfrac aA=\dfrac bB=d\implies(A,B)=1$
$$c^2(A^2+B^2)=A^2B^2$$
$$\implies\left(\dfrac{AB}c\right)^2=A^2+B^2$$ which is an integer
So, $c|AB, c=CAB$(say)
$$\implies A^2+B^2=C^2$$
As $(A,B)=1$
WLOG $A,B\in[m^2-n^2,2mn],C=m^2+n^2$
So, we need $2mn$ to divide $100\iff mn|50$ 
or $m^2-n^2$ to divide $100$
and of course  $(2mn,m^2-n^2)=1$