Find all positive integer solutions to the equation $\frac{1}{a^2}+\frac{1}{b^2}=\frac{1}{c^2}$ 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.

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.
