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.
The slickest way is via a little Algebraic Number Theory. If $p=a^2+b^2=c^2+d^2$ then $$p=(a+bi)(a-bi)=(c+di)(c-di)$$ Now ${\bf Z}[i]$ is a unique factorization domain, so these two factorizations of $p$ show that $a+bi$ can't be a prime in ${\bf Z}[i]$. We must have a non-trivial factorization $a+bi=(s+ti)(u+vi)$, whence $a-bi=(s-ti)(u-vi)$, and then $$p=(s^2+t^2)(u^2+v^2)$$ contradicting primality of $p$.
There are ways to answer your question without these advanced concepts, but I can never remember how it's done. I'm sure someone else will.
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.
The problem is trivialized if you are given that $x^2+y^2=n$ has $r_2(n)=4\sum_{d|n} \sin(\pi d/2) $ solutions in the integers.
The number of ways $r_2(p)=4\sum_{d|p} \sin(\frac{\pi d}{2})=4(1+\sin(\pi p/2))$.
If $\sin(\pi p /2)=1$ then there are 8 solution in the integers. Namely, $(\pm a,\pm b)$ and $(\pm b, \pm a)$ satisfy $x^2+y^2=p$ where $a\neq b$. There is another way one might imagine $8$ solutions: $(0,\pm a),(\pm a,0)$ and $(\pm b,\pm b)$ but this cannot happen because it would mean that $p$ is a perfect square and therefore not prime.
If $\sin(\pi p /2)=-1$ then we cannot write $p$ as the sum of squares.
$If \sin(\pi p /2)=0$ then $p=2$.
This is essentially the same as Gerry Myerson's proof but employing known properties of Gaussian primes:
Suppose $p=a^2+b^2=c^2+d^2$ for some integers $a,b,c,d>0$. Assume, additionally, that $a$ and $c$ are odd whereas $b$ and $d$ are even. We must show $a=c$ and $b=d$.
We know $$p=(a+bi)(a-bi)=(c+di)(c-di),$$ where all the Gaussian integers appearing are primes. Thus, by unique factorization in $\mathbb{Z}[i]$, \begin{equation}a+bi=u\begin{cases}c+di\\ c-di\end{cases},\tag{1}\end{equation} where $u$ is a unit of $\mathbb{Z}[i]$, that is, $u\in \{1,-1,i,-i\}$. Finally, we divide into the four possible cases for $u$:
$\color{blue}{u=1}$. Forces $a+bi=c+di$, and the result follows.
$\color{blue}{u=-1}$. This case is impossible because RHS in $(1)$ would have negative real part, while LHS has positive real part ($a>0$).
$\color{blue}{u=i}$. Then $a+bi=d+ci$ by positiveness, but this contradicts the chosen parity of $a,b,c,d$.
$\color{blue}{u=-i}$. Impossible because RHS in $(1)$ would have negative imaginary part.
If $p=a^2+b^2=c^2+d^2$ then $$p=\frac{(ac+bd)(ac-bd)}{(a+d)(a-d)}$$
