Upsetting Pitagoras

260 /www/nrich/html/content/99/04/15plus1/ 1 2011-02-01T00:00:01 <mdoxml version="1.0"> Find the smallest integer solution to the following equation: $$\frac {1}{x^2}+\frac {1}{y^2}=\frac {1}{z^2}$$ </mdoxml> <mdoxml version="1.0">"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." You can also use <a href="../public/viewer.php?obj_id=1309">Pythagorean Triples</a> to find the smallest integer solution to the equation: 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. </mdoxml> <mdoxml version="1.0">Take a cue from the title! Can you in any way use $a$, $b$ and $c$ where you know $a^2 + b^2 = c^2$? No knowledge is needed here, only mathematical reasoning. It is an important point to make that finding a method for producing solutions does not prove that there is no other method and no smaller solution generated another way. However, once you have found what you believe to be the smallest solution, it is possible to check numerically that there are in fact no smaller solutions.</mdoxml> 2 3 0 0 0 1 1 Upsetting Pitagoras Find the smallest integer solution to the equation: 1/x^2 + 1/y^2 = 1/z^2 Admin Short problems Numbers and the Number System Integers Algebra Diophantine equations 2D Geometry, Shape and Space Pythagorean triples