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

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!

Answer 1

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.

Answer 2

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