Does there exist an $N$ such that $$N=x_1^2+y_1^2=x_2^2+y_2^2=x_3^2+y_3^2\neq x_4^2+y_4^2\qquad \land \qquad \begin{cases}\gcd(x_1,y_1)&=1 \\\gcd(x_2,y_2)&=1\\\gcd(x_3,y_3)&=1 \end{cases}$$ ???

If such a number $N$ is expressible as a sum of two squares in exactly $3$ and only $3$ ways, is it guaranteed that the second condition above FAILs to be satisfied?

My way of deriving such numbers $N$:

$$\text{if}\quad N=\prod_{i=1}^k\underbrace{p_i^{\alpha_i}}_{{\text{all} \quad 1 \pmod 4\quad \text{primes}}}\qquad \text{then} \qquad r(N)=\frac 1 2 \prod_{i=1}^k(\alpha_i+1)$$ with $r(x)$ being the function which is valued by the number of ways a natural number $x$ is expressible as a sum of two squares in distinct ways (without respect to order, or negatives, also not including $0$). For only three ways to express $N$, I chose $k=2$ and let the $\alpha_i$ be $1$ and $2$ only.

Examples, $$\begin{aligned}325&=5^2\cdot 13 &&\to r(325)=3\\ &=1^2+18^2=6^2+17^2=\boxed{10^2+15^2}\\425&=5^2\cdot 17 &&\to r(425)=3 \\&=\boxed{5^2+20^2}=8^2+19^2=13^2+16^2 \\845&=5 \cdot 13^2&&\to r(825)=3 \\&=2^2+29^2=\boxed{13^2+26^2}=19^2+22^2\end{aligned}$$

The below is my first try at a parameterization for the case where $r(N)=3$.

Derivation, in case of interest:

$\begin{cases}x_1&=pq+rs \\ y_1&=pr-qs \\ x_2&= pq-rs &&=ab+cd&&&=u \\ y_2&=pr+qs&&= ac-bd&&&=v\\ x_3 & &&=ab-cd \\ y_3& &&=ac+bd \end{cases} \iff \begin{cases}ab+cd-pq+rs&=0 \\ ac-bd-pr-qs&=0\end{cases}$ $\iff \left[\begin{array}{llll}b&c&-q&r \\ c&-b&-r&-q\end{array}\right]\cdot \left[\begin{array}{l}a \\ d \\ p \\ s\end{array}\right]=\overrightarrow{0}\implies\left[\begin{array}{llll}b&c&-q&r \\ c&-b&-r&-q\end{array}\right]\sim \left[\begin{array}{llll}1 & 0 & -\frac{bq+cr}{b^2+c^2} & \frac{br-cq}{b^2+c^2} \\ 0&1&\frac{br-cq}{b^2+c^2}&\frac{bq+cr}{b^2+c^2}\end{array}\right]\implies$ $a=\left(\frac{bq+cr}{b^2+c^2}\right)p+\left(\frac{cq-br}{b^2+c^2}\right)s=\frac{b(pq-rs)+c(pr+qs)}{b^2+c^2}=\frac{bu+cv}{b^2+c^2}$ $d=\left(\frac{cq-br}{b^2+c^2}\right)p-\left(\frac{bq+cr}{b^2+c^2}\right)s=\frac{-b(pr+qs)+c(pq-rs)}{b^2+c^2}=\frac{cu-bv}{b^2+c^2} \implies$ $x_3=ab-cd=\frac{1}{b^2+c^2}\left[b(bu+cv)-c(cu-bv)\right]=\frac{(b^2-c^2)u+(2bc)v}{b^2+c^2}\sim y_3=\frac{(2bc)u-(b^2-c^2)v}{b^2+c^2}$ $\begin{cases}u&=pq-rs \\ v&=pr+qs\end{cases}\implies \begin{cases}p&=-\frac{qu+rv}{r^2+q^2}\\s&=\frac{ru-qv}{r^2+q^2}\end{cases}$ $\implies x_1=\frac{(r^2-q^2)u-(2qr)v}{r^2+q^2},\qquad y_1=\frac{(2qr)u+(r^2-q^2)v}{r^2+q^2}$

  • 4
    $\begingroup$ as long as the number $N$ is odd and not itself a square, your $r(N)$ is correct : in these cases the total number of representations is a multiple of $8$ . Theorem 65 in Dickson, Introduction to the Theory of Numbers (1929) $\endgroup$
    – Will Jagy
    Dec 16, 2020 at 4:07
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    $\begingroup$ When you write, "proportionality between the two numbers," I think what you mean is that the two numbers are not relatively prime, that is, they have a common divisor other than $1$. Is that right? $\endgroup$ Dec 31, 2020 at 4:11
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    $\begingroup$ And what are your "examples" examples of? They are not examples of $r(n)=3$; are they just examples of computing $r(n)$? $\endgroup$ Dec 31, 2020 at 4:13
  • $\begingroup$ @GerryMyerson yes $\endgroup$ Dec 31, 2020 at 5:03
  • $\begingroup$ @GerryMyerson the examples where $r(N) \neq 3$ have more to do with a previous edit of this question.... $\endgroup$ Dec 31, 2020 at 5:11

1 Answer 1


If $N=p^2q$ where $p\ne q$ are both primes congruent $1\bmod4$, then there exist unique $a,b$ with $p=a^2+b^2$, $a>b>0$, and $c,d$ with $q=c^2+d^2$, $c>d>0$. Then $p^2=(a^2-b^2)^2+(2ab)^2$, and $$\begin{aligned}N&=\left[(a^2-b^2)c+(2ab)d\right]^2+\left[(a^2-b^2)d-(2ab)c\right]^2\\&=\left[(a^2-b^2)d+(2ab)c\right]^2+\left[(a^2-b^2)c-(2ab)d\right]^2\\&=(pc)^2+(pd)^2\end{aligned}$$

Note that $2N$ has the same number of representations as $N$.

Also, if $p$ is a prime congruent $1\bmod4$, then $p^4$ has three representations, e.g., $5^4=625=25^2+0^2=24^2+7^2=20^2+15^2$.

EDIT: Let's look at this over the Gaussian integers, ${\bf G}=\{\,a+bi\mid a,b{\rm\ in\ }{\bf Z}\,\}$. Here are some facts we will use without proof. Proofs can be found all over the web, or in many intro Number Theory and Algebraic Number Theory texts.

  1. $\bf G$ is an integral domain.

  2. The units in $\bf G$ are $\{\,\pm1,\pm i\,\}$. Two elements are called associates if their quotient is a unit. The conjugate of $a+bi$ is $a-bi$. The conjugate of $z$ is denoted $\overline z$.

  3. The primes in $\bf G$ are $1+i$, $a\pm bi$ where $a^2+b^2$ is a $1\bmod4$ prime, each $3\bmod4$ prime, and all their associates.

  4. $\bf G$ is a unique factorization domain. In particular, every $1\bmod4$ prime $p$ has a unique factorization (up to associates) $p=(a+bi)(a-bi)$ where $a^2+b^2=p$.

Now, every expression of $N=r^2+s^2$ as a sum of two squares corresponds to a factorization of $N=(r+si)(r-si)$ as a product of a Gaussian integer and its conjugate. In particular, if $N=p^2q$ where $p=a^2+b^2$ and $q=c^2+d^2$ are distinct $1\bmod4$ primes, then $$N=(a+bi)^2(a-bi)^2(c+di)(c-di)$$ and there are three ways to write this as a product of a Gaussian integer and its conjugate:

  1. $N=[(a+bi)^2(c+di)]\ [(a-bi)^2(c-di)]$

  2. $N=[(a+bi)^2(c-di)]\ [(a-bi)^2(c+di)]$

  3. $N=[(a+bi)(a-bi)(c+di)]\ [(a+bi)(a-bi)(c-di)]$

Multiplying everything out, this gives $N=e\overline e$ where $e$ takes on the values

  1. $(a^2-b^2)c-2abd+((a^2-b^2)d+2abc)i$,

  2. $(a^2-b^2)c+2abd+(-(a^2-b^2)d+2abc)i$,

  3. $pc+pdi$.

And that gives our three ways to write $N$ as a sum of two squares.

It's clear that the representation coming from the third option, $N=(pc)^2+(pd)^2$, involves numbers that are not relatively prime.

  • $\begingroup$ $\gcd(a,b)=\gcd(c,d)=1$. From that, you can show that only the $(pc)^2+(pd)^2$ is not relatively prime. Everything becomes simpler when analyzed in terms of complex numbers (Gaussian integers). We're looking at the different choices of sign in $(a\pm bi)(a\pm bi)(c\pm di)$. If you don't like zero, then $5^5=55^2+10^2=50^2+25^2=41^2+38^2$. $\endgroup$ Dec 31, 2020 at 9:07
  • $\begingroup$ The parametrization may be similar to yours, but you have all these quantities $r,q,u,v,b,c$ with no indication of where they come from, whereas I tie $a,b,c,d$ explicitly to the representations of $p,q$ as sums of two squares. $\endgroup$ Dec 31, 2020 at 9:14
  • $\begingroup$ I've added my derivation to the post. I do like your approach, and your parameterization is likely better, but I just don't see how this parameterization proves that there IS NOT some N with the above second property that every (x,y) pair be mutually prime. Is this the ONLY parameterization possible??? $\endgroup$ Dec 31, 2020 at 16:05
  • $\begingroup$ If $n=p^2q$ then there will be a representation of $n$ obtained by taking the representation of $q=c^2+d^2$ and multiplying both $c$ and $d$ by $p$. So, you can't avoid having a pair that isn't mutually prime. Similar remarks apply to $n=p^5$ and $n=p^6$. $5^6=117^2+44^2=120^2+35^2=100^2+75^2$. $\endgroup$ Dec 31, 2020 at 22:38
  • $\begingroup$ Does that answer your question about relatively prime pairs? $\endgroup$ Jan 2, 2021 at 0:43

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