The expression of a prime of form
\[4k+1\]
as a sum of two squares is unique except for the order of the squares.Proof
Suppose
\[p=a^2+b^2=c^2+d^2\]
with \[a \gt b \gt 0, c \gt d \gt 0\]
.\[a^2d^2-b^2c^2=(p-b^2)d^2-b^2(p-d^2)=p(d^2-b^2) \equiv 0 \; (mod \; p)\]
Then
\[(ad-bc)(ad+bc) \equiv 0 \; (mod \; p)\]
.Euclid's Lemma tells us that either
\[p | (ad-bc)\]
or \[p | (ad+bc)\]
.Suppose
\[p | (ad+bc)\]
. Each of \[a, \; b, \; c, \; d\]
is strictly between 0 \nd \[\sqrt{p}\]
so \[0 \lt ad+bc \lt 2p\]
. It must be the case then that \[ad+bc=p\]
but then\[p^2=(a^2+b^2)(c^2+d^2)=(ad+bc)^2+(ac-bd)^2=p^2+(ac-bd)^2\]
But then
\[ac-bd=0\]
and since \[a \gt b, \; c \gt d\]
we must have \[ac \gt bd\]
so we have a contradiction, and \[p | (ad-bc)\]
.Since
\[0 \lt a, \; b, \; c, \; d \lt \sqrt{p}\]
we have \[=p \lt ad-bc \lt p\]
so \[ad=bc\]
. Then \[a | bc\]
but \[gcd(a,b)=1\]
so \[a | c\]
. Putting \[c=ka\]
the equation \[ad=bc\]
becomes \[ad=bka \rightarrow d=bk\]
and \[p^2=c^2+d^2=(ak)^2+(kb)^2=k^2(a^2+b^2)=k^2p\]
.Hence
\[k=1\]
and \[a=c, \; b=d\]
.