<s>[INST] If a prime can be expressed as sum of square of two integers, then prove that the representation is unique. If a prime can be expressed as sum of two squares, then prove that the representation is unique.  
My attempt:
If $a^2+b^2=p$, then it is obvious that $a,b$ of different parity.
Now, I assume the contraposition that the representation is not unique, $p=a^2+b^2=c^2+d^2$. Again, $c,d$ are of different parity.  
Now, let $b,d$ be even and $a,c$ be odd.  
So, $a^2+b^2=c^2+d^2 \implies a^2-c^2=d^2-b^2 \implies (a+c)(a-c)=(d-b)(d+b)$.  
I cannot proceed any further. Please help. 
 [/INST] Here's an answer without Algebraic Number Theory. I found it in Shanks, Solved and Unsolved Problems in Number Theory. 
Assume $$p=a^2+b^2=c^2+d^2\tag1$$ with all variables positive integers. Then $$p^2=(a^2+b^2)(c^2+d^2)=a^2c^2+a^2d^2+b^2c^2+b^2d^2$$ and you can verify by just multiplying everything out that $$p^2=(ac+bd)^2+(ad-bc)^2\tag2$$ and $$p^2=(ac-bd)^2+(ad+bc)^2\tag3$$ By (1), we have $$(p-a^2)d^2=(p-c^2)b^2$$ which implies $$p(d^2-b^2)=(ad-bc)(ad+bc)\tag4$$ From (4), $p$ divides $ad-bc$, or $p$ divides $ad+bc$. If $p$ divides $ad-bc$, then from (2) we get $ad-bc=0$, so $d^2-b^2=0$, so $b=d$. If $p$ divides $ad+bc$, then from (3) we get $ac=bd$. Now $a$ and $b$ are relatively prime, so $a$ divides $d$, and $b$ divides $c$. Then by (1) we have $a=d$, and we have proved that the two representations of $p$ are the same. 
This is probably something like what @Konstantinos was getting at in his answer. 
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