Upsetting Pythagoras

"Hello. I'm Kim Jinhyuna from Kingston-Grammar School. I would like to inform you that I have worked out the question 'Upsetting Pitagoras'. The mathematical problem has an infinite number of solutions. Let $x=1$ and $y=2$. Then $$\frac{1}{x^2}+\frac{1}{y^2}=\frac{1}{1^2}+\frac{1}{2^2}=1\cdot 25=\frac{1}{0\cdot 8}.$$ So $z^2 = 0{\cdot}8$, and $z = \sqrt{0{\cdot}8}$ which gives $z=\pm 0{\cdot}894$ to 3 significant figures. For another solution, let Let $x=3$ and $y=4$. Then $$\frac{1}{x^2}+\frac{1}{y^2}=\frac{1}{3^2}+\frac{1}{4^2}=1\cdot 736\dots =\frac{1}{5\cdot 76}.$$ So $z^2 = 5{\cdot}76$, and $z = \sqrt{5{\cdot}76}$ which gives $z=\pm 2{\cdot}4$ (exactly). However, I think that the problem comes with the assumption that $x$, $y$ and $z$ are all integers, in which case one answer is $x=30$, $y=40$ and $z=24$; that is $$\frac{1}{{30}^2}+\frac{1}{{40}^2}=\frac{1}{{24}^2}.$$ If we multiply both sides by $4$ we get $$\frac{1}{{15}^2}+\frac{1}{{20}^2}=\frac{1}{{12}^2}.$$ I'm looking forward to more tough and hard questions."

You can also use Pythagorean Triples to find the smallest integer solution to the equation:

If $a^2 + b^2 = c^2$ then $$\frac{1}{b^2{c}^2}+\frac{1}{a^2{c}^2}=\frac{1}{a^2{b}^2}.$$ So every Pythgorean triple gives rise to a solution to our problem. The smallest solution of this type arises from $a=3, b=4, c=5$ and gives $$\frac{1}{{20}^2}+\frac{1}{{15}^2}=\frac{1}{{12}^2}.$$ Notice we have not proved that there is no other way of producing solutions but a simple computer program to make an exhaustive check of values of $x, y$ and $z$ up to these values will prove that there are no smaller solutions.