Maths challenge 1
Posted by Chris on April 28, 2011 – 2:27 pm
Solve x² + y² + z² = 2xyz, where x,y,z are integers > 0 (i.e. natural numbers).
Addendum: In fact that isn’t possible to do. So I’ve ended up proving that
x² + y² + z² = 2n xyz cannot be solved for n,x,y,z in natural numbers.