The number of ways of writing an integer as a sum of two squares

Question

Given an integer $m=pq$, where $p,q$ are both primes such that $p\equiv 1 \pmod{4}, q\equiv 1 \pmod{4}$. It is known that $p$ can be written as a sum of two squares (of positive integers) in a unique way, and the same for $q$. Prove that $m$ can be written as a sum of two squares (of positive integers) in exactly two distinct ways.

Attempt

Notice that for positive integers $u,v,A,B$, we have \begin{align*}(u^2+v^2)(A^2+B^2)&=(uA+vB)^2+(vA-uB)^2=(vA+uB)^2+(uA-vB)^2\end{align*} Therefore if $p=a^2+b^2,q=c^2+d^2$, then \begin{align*}m&=(a^2+b^2)(c^2+d^2)=(ac+bd)^2+(bc-ad)^2=(bc+ad)^2+(ac-bd)^2\end{align*} meaning that $m$ can be written as a sum of two squares (of positive integers) in at least two distinct ways.

How do I prove that there doesn't exist a third way of writing $m$ as a sum of two squares (of positive integers)?

Answer 1

This is strictly related with the number of ways of writing $pq$ as $z\bar{z}$ in $\mathbb{Z}[i]$, that is an Euclidean domain, hence a UFD. See this other question for details.

@LiebsterJugendtraum: my answer involves ring theory, more than group theory, but just a little. To this "by hand", you can use Lagrange's identity and try to prove that, assuming that $pq$ has $3$ (or more) distinct representations as a sum of two squares, then $(-1)$ is the square of too many elements in $\mathbb{F}_p^*$ or in $\mathbb{F}_q^*$.

Answer 2

I think we can always translate everything to the basics operations but it will take a huge number of pages! Fortunately, your question has an elementary answer. We denote by $S_n$ the set of representation of the integer $n$ as sum of squares, we have $(a,b)\in S_n$ if and only if $0\leq b<a$ and $ a^2+b^2=n$;

Given two primes $p$ and $q$ let: $$ S_p=\{(a_p,b_p)\}\\ S_q=\{(a_q,b_q)\} $$ Now we want to prove that $S_{pq}$ contains exactly two elements, or in other words we want to prove that the following equations on unknown $(a,b),(c,d)$ has exactly one solution :

• $a>b>0$ and $c>d> 0$ (it's clear that neither $d=0$ or $b=0$ occurs)
• $a^2+b^2=c^2+d^2=pq$
• $a>c$ ( different elements)

The equation in the middle is equivalent to: $$(a-c)(a+c)=(d-b)(d+b)$$ this implies the existence of integers $x,y,z,t$ pairwise coprime such that: $$\begin{align} a-c&=&2xy\\ a+c&=&2zt\\d-b&=&2xz\\d+b&=&2yt \end{align}$$ Note that the existence is not hard for example $x=gcd\big(\frac{a-c}{2},\frac{d+b}{2}\big),y=gcd\big(\frac{a-c}{2},\frac{d-b}{2}\big)\cdots$, so this will justify also that they are pairwise corprime.

The result is the fact that: $$pq=(x^2+t^2)(y^2+z^2)$$ obviously $xyzt\neq 0$ therefore $\{p,q\}=\{x^2+t^2,y^2+z^2\}$ we have two cases:

1. $p=x^2+t^2$ and $q=y^2+z^2$, because $a+c>d-b$ we have $t>x$, and $a+c > d+b$ implies $z>y$ so $x=b_p,t=a_p,y=b_q, z=a_q$ so $a=a_pa_q+b_qb_p,b=a_pb_q-b_pa_q,c=a_pa_q-b_qb_p, d=a_pb_q+b_pa_q$
2. $q=x^2+t^2$ and $p=y^2+z^2$ because of the symetry it gives different $x,y,z,t$ but the same $a,b,c,d$ as the first case.

So there is only one solution to the equations hence $S_{pq}$ contains exactly two elements.

Answer 3

@Elaqqad

This is my solution based on your idea and notations. Please point out wherever there is a mistake.

There exist coprime integers $x,y,z,t$ such that

\begin{align*} &a-c=xy\\& a+c=zt\\& d-b=xz\\& d+b=yt \end{align*}

Therefore we have that

\begin{align*} &a=\frac{1}{2}(xy+zt)\\& c=\frac{1}{2}(zt-xy)\\& d=\frac{1}{2}(xz+yt)\\& b=\frac{1}{2}(yt-xz) \end{align*}

Consequently, we obtain that

\begin{align*} &a^2+c^2=\frac{1}{2}(x^2y^2+z^2t^2)\\& b^2+d^2=\frac{1}{2}(x^2z^2+y^2t^2) \end{align*}

which imply that

\begin{align*} &2pq=(a^2+b^2)+(c^2+d^2)=(a^2+c^2)+(b^2+d^2)=\frac{1}{2}(x^2+t^2)(y^2+z^2) \end{align*}

So we have that

\begin{align*}pq=\frac{1}{4}(x^2+t^2)(y^2+z^2)\end{align*}

meaning that there are three possible cases:

\begin{align*} &\{p,q\}=\{\frac{1}{2}(x^2+t^2),\frac{1}{2}(y^2+z^2)\}\\& \{p,q\}=\{\frac{1}{4}(x^2+t^2),(y^2+z^2)\}\\& \{p,q\}=\{(x^2+t^2),\frac{1}{4}(y^2+z^2)\} \end{align*}

Case 1.$\{p,q\}=\{\frac{1}{2}(x^2+t^2),\frac{1}{2}(y^2+z^2)\}$:

If a prime $p_0$ satisfies that $p_0\equiv 1 \pmod 4$, then $p_0=m^2+n^2$ for some positive integers $m,n$ with different parity.

Using $(u^2+v^2)(A^2+B^2)=(uA+vB)^2+(vA-uB)^2=(vA+uB)^2+(uA-vB)^2$, we obtain that $2p_0=(1^2+1^2)(m^2+n^2)=(m+n)^2+(m-n)^2$. The right-hand side is the unique way of writing $2p_0$ as a sum of two squares.

Up to permutation, we may assume that $p=\frac{1}{2}(x^2+t^2),q=\frac{1}{2}(y^2+z^2)$.

If $p=u^2+v^2,q=A^2+B^2$, where $u>v,A>B$,and $u>A)$, then we obtain that

$t>x,z>y$, so $t=u+v,x=u-v,z=A+B,y=A-B$.

\begin{align*} &a=\frac{1}{2}(xy+zt)=uA+vB\\& c=\frac{1}{2}(zt-xy)=vA+uB\\& d=\frac{1}{2}(xz+yt)=uA-vB\\& b=\frac{1}{2}(yt-xz)=vA-uB \end{align*}

Case 2.$\{p,q\}=\{\frac{1}{4}(x^2+t^2),(y^2+z^2)\}$:

If a prime $q_0$ satisfies that $q_0\equiv 1 \pmod 4$, then $q_0=m^2+n^2$ for some positive integers $m,n$ with different parity.

Likewise, we obtain that $4q_0=(0^2+2^2)(m^2+n^2)=(2m)^2+(2n)^2$. The right-hand side is the unique way of writing $4q_0$ as a sum of two squares.

Using the same notations, we have that $t=2u,x=2v,z=A,y=B$, then

\begin{align*} &a=\frac{1}{2}(xy+zt)=vB+uA\\& c=\frac{1}{2}(zt-xy)=uA-vB\\& d=\frac{1}{2}(xz+yt)=vA+uB\\& b=\frac{1}{2}(yt-xz)=uB-vA \end{align*}

Case 3.$\{p,q\}=\{\frac{1}{4}(x^2+t^2),(y^2+z^2)\}$:

This is analogous to case 2.

Consequently, we know that $\{(a,b),(c,d)\}=\{(uA+vB,vA-uB),(vA+uB,uA-vB)\}$, so the product of two distinct primes can be written as a sum of two squares in exactly two ways. $\blacksquare$