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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
\[\begin{equation} \begin{aligned} 2p &= (1^1+1^2)(a^2+b^2) \\ &= (1 \times a+1 \times b)^2+(1 \times a-1 \times b)^2 \\ &= (a+b)^2+(a-b)^2 \end{aligned} \end{equation}\]
.
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).