2 to even power as a sum of squares

Problem: Show that if $\displaystyle 2^{2k} = a^{2} + b^{2}$ 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 $\displaystyle (a, b, 2^{k}) $

> Show that we cannot obtain a primitive P.T. from above ($\displaystyle 2^{k}$ is not odd) so the equation $\displaystyle (2^{k})^{2} = a^{2} + b^{2}$ does not have a solution for any k

> So, we cannot have a>0, b>0

> Show that $\displaystyle 2^{2k} = (2^{k})^{2} + 0^{2}$ so (a>0 and b=0) OR (a=0 and b>0)

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

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=2^{k}, and similarly if b=0