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: 
$$1^2+2^2=5\cdot 2^0$$
$$1^2+3^2=5\cdot 2^1$$
$$2^2+4^2=5\cdot 2^2$$
$$2^2+6^2=5\cdot 2^3$$

Note that $5$ is a sum of two squares, for $5=2^2+1^2$.
If $n$ is even, say $n=2k$, then $5\cdot 2^n=(2^k\cdot 2)^2+(2^k\cdot 1)^2$.
Also, $10=3^2+1^2$. So if $n$ is odd, say $n=2k+1$, then $5\cdot 2^{n}=10\cdot 2^{2k}=(2^k\cdot 3)^2+(2^k\cdot 1)^2$.
