"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 $${1\over x^2}+{1\over y^2}= {1\over 1^2}+{1\over 2^2} = 1{\cdot}25 = {1\over 0{\cdot}8}.$$ 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 $${1\over x^2}+{1\over y^2}= {1\over 3^2}+{1\over 4^2} = 1{\cdot}736\ldots = {1\over 5{\cdot}76}.$$ 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 $${1\over 30^2}+{1\over 40^2} = {1\over 24^2}.$$ If we multiply both sides by $4$ we get $${1\over 15^2}+{1\over 20^2} = {1\over 12^2}.$$ I'm looking forward to more tough and hard questions."
If $a^2 + b^2 = c^2$ then $${1\over b^2c^2} + {1\over a^2c^2} = {1\over a^2b^2}.$$ 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 $${1\over 20^2} + {1\over 15^2} = {1\over 12^2}.$$ 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.
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