\[p=4k+1\]
is prime then \[2p\]
can be written as the sum of two square uniquely neglecting the order in which they are written.Any prime of the form
\[4k+1\]
can be written as the sum of two squares \[p=a^2+b^2\]
using Primes of Form 4k+1 as Sum of Two Squares.Now use
\[\begin{equation} \begin{aligned} 2p &= (1^1+1^2)(a^2+b^2) \\ &= (1 \times a+1 \times b)^2+(1 \times a-1 \times b)^2 \\ &= (a+b)^2+(a-b)^2 \end{aligned} \end{equation}\]
.To prove uniqueness suppose
\[2p=x^2+y^2\]
(1) then \[x^2+y^2 \equiv 2 \; (mod \; 4)\]
so \[x, \; y\]
must both be odd.\[2p =x^2+y^2=2(( \frac{x+y}{2})^2+( \frac{x-y}{2})^2)\]
so we can also have \[p=( \frac{x+y}{2})^2+( \frac{x-y}{2})^2\]
but from Uniqueness of Expressing Primes of Form 4k+1 as Sum of Two Squares this expression is unique, therefore so is (1).