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Diophantine equation: $\frac{1}{a^2}+\frac{1}{b^2} =\frac{1}{c^2}$ What is sum of all a equal or less than $100$?

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    What exactly are you asking? Your question is completely unclear. – 2017-02-06
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    The problem is from AOPS AMC 10 practice test problem 25. a, b,c are positive integers to the the equation. Find all solution of 'a' and sum them up. a is less or equal than 100. – 2017-02-06
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    As seen [here](https://en.wikipedia.org/wiki/Fermat%27s_Last_Theorem#n_.3D.C2.A0.E2.80.932) primitive solutions are given. – 2017-02-06
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    http://math.stackexchange.com/questions/1634944/integer-solutions-for-frac1x2-frac1y2-frac1z2/1635514#1635514 – 2017-02-06

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Since it's diophantine, I assume you want $a,b,c$ to be positive integers. Write it as $$c = \frac{ab}{\sqrt{a^2+b^2}}$$

Let $g = \gcd(a,b)$, so $a = ug$ and $b = vg$ with $\gcd(u,v)=1$.
Then $a^2 + b^2 = (u^2 + v^2) g^2$, so the equation becomes

$$ c = \frac{uvg^2}{g \sqrt{u^2+v^2}} = \frac{uvg}{\sqrt{u^2+v^2}}$$ Any prime that divides $u^2+v^2$ can't divide $u$ (because then it wouldn't divide $v$), and similarly can't divide $v$, so must divide $g$. Thus $\sqrt{u^2+v^2}$ must divide $g$, while $(u,v, w=\sqrt{u^2+v^2})$ must be a primitive Pythagorean triple.

The primitive Pythagorean triples can be parametrized by

$$ u =2 m n,\; v = m^2 - n^2,\; w = m^2 + n^2$$ with $1 \le n < m$, $m$ and $n$ coprime and not both odd.

You want $a \le 100$ so $u \le 100$, then $mn \le 50$. Not too many possibilities ...