The number of the solutions $(x_0,y_0)$ for $x^2+y^2=z_0^2$ such that $(x_0,y_0) \in \mathbb{N} \times \mathbb{N}$

Question

Let $\mathbb{N}=\{1,2,3,...\}$ be the set of natural numbers. (We assume that the zero does not belong to $\mathbb{N})$.

Suppose we have the following equation: $$x^2+y^2=z_0^2$$ where $z_0 \in \mathbb{N}$ is given.

We are interested in finding the number of the solutions $(x_0,y_0)$ for the equation above such that $(x_0,y_0) \in \mathbb{N} \times \mathbb{N}$.

Two points should be taken into consideration:

1. The equation may have no solution (for instance, take $z_0=1$).
2. If $(a,b) \in \mathbb{N} \times \mathbb{N}$ is a solution then $(b,a) \in \mathbb{N} \times \mathbb{N}$ is obviously a solution too. Therefore, we can consider $(a,b)$ and $(b,a)$ as the same solution.

Are there articles, books, ideas or anything else about this idea?

Answer 1

There is a fairly simple expression for all representations $x^2 + y^2 = n$ when we can factor$n.$ If $n$ is a square, there are exactly four of these with a $0.$ For you, $n$ is never twice a square, so getting $1 \leq x < y$ comes from counting the nonzero representations and dividing that by $8$

This book may be read online, probably downloaded

Answer 2

There are many books relating to the Pythagorean theorem $(A^2+B^2=C^2)\,$ and the special case of Pythagorean triples where $A,B,C\in\mathbb{N}.$

Some of the books in my library are

1. PYTHAGORAS' THEOREM The Sacred Geometry of Triangles by Claudi Alsina

2. The Pythagorean Triangle: Or the Science of Numbers by George Oliver

3. A Pythagorean Introduction to Number Theory: Right Triangles, Sums of Squares, and Arithmetic (Undergraduate Texts in Mathematics) by Ramin Takloo-Bighash

4. Pythagorean triangles by Waclaw SIERPINSKI

You may find others in your local library but here is a summary of the little you might glean from them to answer your question.

There may be primitive and/or imprimitive solutions to $\space a^2+b^2= z_o^2.\quad $ All valid $z_0$-values are of the form $4k+1$ but not all $4k+1$ values are valid. $\space$ A list of valid primitive $z_0$-values is shown in the Online Encyclopedia of Integer Sequences. $\space$ If a number is shown more than once, it means there is more than one solution for that number.

The number of primitive solutions for any given $z_0$ is $2^{n-1}$ where $n$ is the number of distinct prime factors of $z_0.\space$ For example, $\space $ given $65=5\times13,\space$ there are $2^{2-1}=2\space$ solutions: $\quad(33,56,65)\qquad (63,16,65).$

Whereas, $z_0=125=5\times5\times5\space$ has only $2^{1-1}=2^0=1\space $ primitive solution. There are $\quad (75,100,125)\space\text{ and }\space (117,44,125)\quad $ but the first is imprimitive.

For solutions given $z_0$ we use a formula with $(x,y,z)=(A,B,C),\space$ one which generates a valid triple for every pair of natural numbers $(m>k).$

$$ \qquad A=m^2-k^2\qquad B=2mk \qquad C=m^2+k^2\qquad$$

We now solve the $C$-function for $k$ and test a defined range of $m$-values to see which, if any, yield integers. We demo with $\quad C=1105=5\times13\times17\space \text{suggesting}\space 2^{3-1}=4\quad $ solutions.

\begin{equation} C=m^2+k^2\implies k=\sqrt{C-m^2}\\ \text{for}\qquad \bigg\lfloor\frac{ 1+\sqrt{2C-1}}{2}\bigg\rfloor \le m \le \big\lfloor\sqrt{C-1}\space\big\rfloor \end{equation}

The lower limit ensures $m>k$ and the upper limit ensures $k\in\mathbb{N}$.

$$C=1105\implies \bigg\lfloor\frac{ 1+\sqrt{2*1105-1}}{2}\bigg\rfloor=24 \le m \le \big\lfloor\sqrt{1105-1}\space\big\rfloor=33\\ \land \quad m\in\big\{24,31,32,33\big\}\implies k\in\big\{23,12,9,4\big\}\\$$

$$ f(24,23)=(47,1104,1105)\quad f(31,12)=(817,744,1105)\\ f(32,9)=(943,576,1105)\quad f(33,4)=(1073,264,1105)\ $$