# 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 ...

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

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.

Hint 1) $$a$$ must be a multiple of $$5$$.

Hint 2) You need only consider multiples of the $$(3,4,5)$$ and $$(5,12,13)$$ triangles.

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 $$(2mn,m^2-n^2)=1$$