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Find all positive integers to the equation $\frac{1}{a^2}+\frac{1}{b^2}=\frac{1}{c^2}$ Multiply both sides with $(abc)^2$ to get $(bc)^2 + (ac)^2 = (ab)^2$.

I then tried some pythagorean triples and nothing worked so I assumed that there arent any solutions. Thank you for your responses.

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    can you fix your $$\LaTeX$$ please? – 2017-01-06
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    This doesn't answer the "*all*" part of the question, but note that every pythagorean triple $u^2+v^2=w^2$ gives a solution $a=uw, b=vw, c=uv\,$. For example $3,4,5$ gives $a=15,b=20,c=12\,$. – 2017-01-06
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    So the equation has infinetaly makt. – 2017-01-06

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Rewrite the equation as $$ \frac{c^2}{a^2}+\frac{c^2}{b^2}=1 $$ and set $c/a=x/z$, $c/b=y/z$, where $x$ and $y$ are coprime.

Then $(x,y,z)$ is a primitive Pythagorean triple, so, without loss of generality, $$ x=u^2-v^2,\quad y=2uv,\quad z=u^2+v^2 $$ where $\gcd(u,v)=1$, one among $u$ and $v$ being even.

Thus $$ \frac{c}{a}=\frac{u^2-v^2}{u^2+v^2},\quad \frac{c}{b}=\frac{2uv}{u^2+v^2} $$ Therefore, for some positive integers $h$ and $k$, $$ c=h(u^2-v^2)=2kuv,\quad a=h(u^2+v^2),\quad b=k(u^2+v^2) $$ Note however that $u^2-v^2$ and $2uv$ are coprime, so $h=2muv$ and $k=m(u^2-v^2)$, so $$ a=2muv(u^2+v^2),\quad b=m(u^4-v^4),\quad c=2muv(u^2-v^2) $$

It's easy to see that any choice of $u$ and $v$ as before and any $m$ produces a solution of the original equation.

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As you noted $1/a^2+1/b^2=1/c^2 \iff b^2+a^2=\frac{b^2a^2}{c^2}$.

This implies that $a$ and $b$ are part of a pythagorean triple and can hence be written $a=k(m^2-n^2)$ and $b=2kmn$.

We therefore have $a^2+b^2=k^2(m^4-2m^2n^2-n^4+4m^2n^2)=k^2(m^2+n^2)^2$

So we need for $k^2(m^2+n^2)^2$ to divide $2k^4(m^2-n^2)m^2n^2$.

The number of solutions is clearly infinite, for every $m$ and $n$ we have an infinite number of solutions, we just need to take $k$ so that $(m^2+n^2)^2| 2k^2(m^2-n^2)m^2n^2$.