Sum of squares and $5\cdot2^n$ Does anyone know of a proof of the result that $5\cdot2^n$ where $n$ is a nonnegative integer is always the sum of two squares?
That is, nonzero integers $x,y$ must always exist where: 
$x^2+y^2=5\cdot2^n$
when $n$ is $0$ or a positive integer?
For example: 
$$1^2+2^2=5\cdot 2^0$$
$$1^2+3^2=5\cdot 2^1$$
$$2^2+4^2=5\cdot 2^2$$
$$2^2+6^2=5\cdot 2^3$$

Consider the following two cases:
$$n=2k\implies 5\cdot 2^{2k}=(1+4)\cdot 2^{2k}=2^{2k}+2^{2k+2}=\left(2^{k}\right)^2+\left(2^{k+1}\right)^2$$
$$n=2k+1\implies 5\cdot 2^{2k+1}=10\cdot 2^{2k}=(1+9)\cdot 2^{2k}=2^{2k}+9\cdot 2^{2k}=\left(2^{k}\right)^2+\left(3\cdot 2^{k}\right)^2$$
