Integer solutions to x² + y² = z

Revision en7, by adamant, 2023-05-16 15:54:58

Hi everyone!

As I continue working through Project Euler, I want to write a blog about another piece of general mathematical knowledge that is both interesting on its own and might be useful in some problems. Consider the following Diophantine equation:

$$$ x^2 + y^2 = z. $$$

We assume that we're given a specific number $$$z \in \mathbb Z$$$, and we need to check if there are $$$x, y \in \mathbb Z$$$ for which the identity above holds. Then, if such numbers exist we should find them and report. Example of a problem that might need it:

Timus — 1593. Square Country. Version 2. You're given an integer $$$n$$$. Find minimum $$$k$$$ such that $$$n = a_1^2+\dots+a_k^2$$$.

Tl;dr.

Let $$$z = 2^{k_0} p_1^{k_1} \dots p_n^{k_n} p_{n+1}^{k_{n+1}} \dots p_m^{k_m}$$$, where $$$p_1, \dots, p_n$$$ are different prime numbers with remainder $$$3$$$ modulo $$$4$$$, and $$$p_{n+1}, \dots, p_m$$$ are different prime numbers with remainder $$$1$$$ modulo $$$4$$$. Then there are two cases. If any of $$$k_{1}, \dots, k_n$$$ is odd, there are no solutions. Otherwise there is always a solution $$$z = x^2 + y^2$$$ that looks like

$$$ x+ iy = (1+i)^{k_0} p_{1}^{k_{1}/2} \dots p_n^{k_n/2} (x_{n+1}+iy_{n+1})^{k_{n+1}} \dots (x_m + iy_m)^{k_m}, $$$

where $$$i^2=-1$$$ and $$$x_k^2+y_k^2 = p_k^2$$$ for $$$k$$$ from $$$n+1$$$ to $$$m$$$. For each $$$p_k$$$, to find such $$$x_k, y_k$$$ we need to find an integer $$$i$$$ such that $$$i^2 \equiv -1 \pmod{p}$$$, then find a minimum $$$x_k = i y_k \bmod p_k$$$ for $$$1 \leq y_k < \sqrt {p_k}$$$. This is doable in $$$O(\log p_k)$$$.

And if we want to count solutions, their number is given by Jakobi's two-square theorem: The number of ways of representing $$$z$$$ as the sum of two squares is $$$4(d_1(z) - d_3(z))$$$, where $$$d_k(z)$$$ is the number of divisors of $$$z$$$ that have remainder $$$k$$$ modulo $$$4$$$.


From exact equation to $$$\bmod z$$$

First of all, let's solve a bit easier problem. One obvious thing we can do is to take remainder modulo $$$z$$$ on both parts:

$$$ x^2+y^2 \equiv 0 \pmod z. $$$

This relaxed version is equivalent to finding $$$x, y, k \in \mathbb Z$$$ such that

$$$ x^2 + y^2 = kz. $$$

Hmmm... Remainders modulo arbitrary number $$$z$$$ is not the most pleasant thing to work on directly. But remainders modulo prime number $$$p$$$ are usually nice. On the other hand, if $$$p$$$ is some prime factor of $$$z$$$ and there is a solution for $$$z$$$, it means that there will as well be a solution for $$$p$$$ with $$$k=\frac{z}{p}$$$. So, let's assume $$$z$$$ to be prime, for now.

From arbitrary $$$z$$$ to prime $$$p$$$

Now, we have another equation

$$$ x^2 + y^2 \equiv 0 \pmod{p}, $$$

where $$$p$$$ is a prime number. What's good about prime numbers is that remainders modulo prime numbers form a field (i.e. they work very similarly to rationals, and we can expect similar results to hold). For $$$p=2$$$, there is a non-trivial solution $$$x=y=1$$$. What about odd numbers $$$p$$$? There are two cases to consider, as the remainder of $$$p$$$ modulo $$$4$$$ is either $$$1$$$ or $$$-1$$$.

Fermat's theorem on sums of two squares tells us that for an odd prime $$$p$$$, the solution exists if and only if $$$p$$$ has a remainder $$$1$$$ modulo $$$4$$$. Moreover, the sum of two squares theorem tells us that the number $$$z$$$ is expressible as a sum of two squares if and only if its prime decomposition does not have a term $$$p^k$$$, where $$$p \equiv -1 \pmod 4$$$, and $$$k$$$ is odd. Let's find out why.

$$$p \equiv 1 \pmod 4$$$

Of course, it's not yet clear why these two cases are important. Let's assume that there is an integer $$$i$$$ such that

$$$ i^2 \equiv -1 \pmod{p}, $$$

that is there is a remainder modulo $$$p$$$ which behaves very similarly to imaginary unit from complex numbers. Then

$$$ x^2 + y^2 \equiv (x+iy)(x-iy) \pmod p. $$$

This reduces the initial equation to a union of linear equations

$$$ \left[ \begin{array}{ll} x+iy \equiv 0 \pmod p, \\ x-iy \equiv 0 \pmod p. \end{array} \right . $$$

For each $$$y$$$, except $$$y=0$$$, there are $$$2$$$ possible values of $$$x = \pm iy$$$, so there are a total of $$$2p+1$$$ solutions. Noteworthy, it is always possible to find a pair of solutions $$$(x,y)$$$ such that $$$1 \leq x, y < \sqrt p$$$, which means that $$$x^2 + y^2 = p$$$ is satisfied exactly.

How to find it? Find $$$i$$$, and consider the minimum value of $$$ik\bmod p$$$ among $$$1 \leq k < \sqrt p$$$. Due to pigeonhole principle, there will be $$$k_1 \neq k_2$$$ such that $$$i (k_1 - k_2) \bmod p \leq \sqrt p$$$. This is actually very similar to 102354I - From Modular to Rational!

Now, when does such $$$i$$$ exist and how to find it? It is known that remainders modulo $$$p$$$ have a primitive root $$$g$$$ such that its powers from $$$0$$$ to $$$p-2$$$ run through all possible remainders modulo $$$p$$$. Note that for odd $$$p$$$ it always holds that

$$$ -1 \equiv g^{\frac{p-1}{2}} \pmod p. $$$

Then, if such $$$i$$$ exists we should be able to find it from

$$$ i^2 \equiv g^{\frac{p-1}{2}} \pmod p \iff i \equiv g^{\frac{p-1}{4}} \pmod p. $$$

Well, technically $$$-g^{\frac{p-1}{4}}$$$ also can be used as $$$i$$$, but it's not that important. What's important is that it is possible to do as above only when $$$p-1$$$ is divisible by $$$4$$$. In other words, when $$$p \equiv 1 \pmod 4$$$.

$$$p \equiv -1 \pmod 4$$$

Now, let's think about the other case. If there is no such $$$i$$$, we can introduce it! Now, we can formally consider numbers that look like $$$x+iy$$$, where $$$i$$$ is not a remainder modulo $$$p$$$. Numbers of this kind, if treated formally, also form a field. If you're familiar with field theory, I should mention that it is isomorphic to the Galois field $$$GF(p^2)$$$. If you're not familiar with it, ignore what I just wrote.

The thing now is that we can try to find all solutions in this new, extended field. And it reduces to the same union of equations

$$$ \left[ \begin{array}{ll} x+iy \equiv 0 \pmod p, \\ x-iy \equiv 0 \pmod p, \end{array} \right . $$$

so for every $$$y$$$, the only possible solutions are $$$x = \pm iy$$$. The problem is, this time such $$$x$$$ would not be a remainder modulo $$$p$$$, unless $$$y=0$$$. Instead, it will be an "imaginary" solution. So, the only "real" solution is $$$x \equiv y \equiv 0 \pmod p$$$. It means that all solutions to

$$$ x^2 + y^2 = kp $$$

look like $$$x = px'$$$ and $$$y=py'$$$. Thus,

$$$ p^2 x'^2 + p^2 y'^2 = kp. $$$

So, if $$$k$$$ is not divisible by $$$p$$$, there are no solutions. Otherwise $$$k=pk'$$$ reduces it to

$$$ x'^2+y'^2 = k', $$$

after which similar argument could be applied. So, if $$$k'$$$ is divisible by an odd power of such $$$p$$$, there are no solutions. We're only one step away from solving the whole $$$x^2+y^2=z$$$ problem now, assuming that we know the factorization of $$$z$$$.

Back to arbitrary $$$z$$$

Now we need to use one more fact from complex numbers. There, we can introduce a norm

$$$ \|x+iy \| = (x+iy)(x-iy) = x^2+y^2. $$$

Its crucial property for this task is that it is multiplicative, that is

$$$ \|(a+ib)(c+id)\| = \| a + ib \| \cdot \| c + id \| . $$$

This gives the Brahmagupta–Fibonacci identity

$$$ (a^2+b^2)(c^2+d^2)= (ac-bd)^2 + (ad+bc)^2, $$$

from which it follows that if we can represent $$$z$$$ as a product of several several numbers that are expressible as a sum of two squares, we can use the identity above to also express $$$z$$$ as a sum of two squares. In complex number terms, it means that we will find a complex number $$$x+iy$$$ such that $$$\|x+iy\| = z$$$.

Repeating what's written in tl'dr section, let $$$z = 2^{k_0} p_1^{k_1} \dots p_n^{k_n} p_{n+1}^{k_{n+1}} \dots p_m^{k_m}$$$, where $$$p_1, \dots, p_n$$$ are different prime numbers with remainder $$$3$$$ modulo $$$4$$$, and $$$p_{n+1}, \dots, p_m$$$ are different prime numbers with remainder $$$1$$$ modulo $$$4$$$. Then there are two cases. If any of $$$k_{1}, \dots, k_n$$$ is odd, there are no solutions. Otherwise there is always a solution $$$z = x^2 + y^2$$$ that looks like

$$$ x+ iy = (1+i)^{k_0} p_{1}^{k_{1}/2} \dots p_n^{k_n/2} (x_{n+1}+iy_{n+1})^{k_{n+1}} \dots (x_m + iy_m)^{k_m}, $$$

where $$$i^2=-1$$$ and $$$x_k^2+y_k^2 = p_k^2$$$ for $$$k$$$ from $$$n+1$$$ to $$$m$$$. This result is tightly connected to the following ones:

Classification of Gaussian primes. A Gaussian integer $$$a+ib$$$ is prime if and only if either of the following is true:

  • $$$a=0$$$ or $$$b=0$$$ and the non-zero number is prime with remainder $$$3$$$ modulo $$$4$$$,
  • $$$a^2 + b^2$$$ is $$$2$$$ or a prime number with remainder $$$1$$$ modulo $$$4$$$.

The result also allows to factor Gaussian integers by representing their norm as a sum of two squares in the way suggested above.

Jakobi's two-square theorem. The number of ways of representing $$$z$$$ as the sum of two squares is $$$4(d_1(z) - d_3(z))$$$, where $$$d_k(z)$$$ is the number of divisors of $$$z$$$ that have remainder $$$k$$$ modulo $$$4$$$.

If there is a prime divisor with remainder $$$3$$$ mod $$$4$$$, it's $$$0$$$. Otherwise, it is $$$4$$$ times the number of divisors of $$$p_{n+1}^{k_{n+1}} \dots p_m^{k_m}$$$. We may interpret it that the divisor decides how much of multipliers in the product would correspond to $$$x_{k} + iy_k$$$, and how much to $$$y_k + i x_k$$$, after which $$$4$$$ accounts for all the possible ways to multiply $$$x+iy$$$ with $$$1$$$, $$$-1$$$, $$$i$$$ or $$$-i$$$, which accounts for all possible combinations of $$$(\pm x, \pm y)$$$.

And, of course, there are similar results for $$$3$$$ squares and $$$4$$$ squares:

Tags number theory, tutorial

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en1 English adamant 2023-05-16 15:26:20 9615 Initial revision (published)