## For p=4k+1 Prime, Writing 2p Uniquely as Sum of Two Squares

Is
$p=4k+1$
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{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}
.
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).