# Math Help - 2 to even power as a sum of squares

1. ## 2 to even power as a sum of squares

Problem: Show that if $2^{2k} = a^{2} + b^{2}$ then either a or b is zero for some integers a, b and a non-negative integer k.

My solution attempt outline:

> Assume a>0, b>0
> Consider a Pythagorean Triple of form $(a, b, 2^{k})$
> Show that we cannot obtain a primitive P.T. from above ( $2^{k}$ is not odd) so the equation $(2^{k})^{2} = a^{2} + b^{2}$ does not have a solution for any k
> So, we cannot have a>0, b>0
> Show that $2^{2k} = (2^{k})^{2} + 0^{2}$ so (a>0 and b=0) OR (a=0 and b>0)

Anything wrong with this? Does anyone have a simpler solution? Thanks.

2. ## Re: 2 to even power as a sum of squares

Looks good.

Last line is a bit confusing. I'd just say that if a=0 then clearly b=2k, and similarly if b=0