[Math] Find all triples of positive integers (x,y,z) such that

Question

Find all triples of positive integers (x,y,z) such that

$x^{z+1} \ – \ y^{z+1}=2^{100}$

The RHS is even, then x and y must be odd and $x^{z+1}>y^{z+1}$, but how to find out them all ?

Answer 1

You can factor $x^{z+1}-y^{z+1}$ by dividing out $x-y$. This shows that $x-y$ is a power of $2.$ But they must only have the same parity-they can both be even. In particular, there are many solutions with $z=0$ If $z=1$ you can factor it as $(x-y)(x+y)=2^{100}$ and you can find a finite number of solutions here. I think, but have not proved, that you will not find any with larger $z$.