If a prime can be expressed as sum of square of two integers, then prove that the representation is unique.

Question

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.

Answer 1

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.

Answer 2

If $p=a^2+b^2=c^2+d^2$ then $$p=\frac{(ac+bd)(ac-bd)}{(a+d)(a-d)}$$

Answer 3

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.

Answer 4

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 $u$:

1. $\color{blue}{u=1}$. Forces $a+bi=c+di$, and the result follows.

2. $\color{blue}{u=-1}$. This case is impossible because RHS in $(1)$ would have negative real part, while LHS has positive real part ($a>0$).

3. $\color{blue}{u=i}$. Then $a+bi=d+ci$ by positiveness, but this contradicts the chosen parity of $a,b,c,d$.

4. $\color{blue}{u=-i}$. Impossible because RHS in $(1)$ would have negative imaginary part.