# Finding Solutions to a System of Diophantine Equations

I'm trying to find triplets of integer $$(x, y)$$ pairs - $$(x_1, y_1), (x_2, y_2), (x_3, y_3)$$ - that satisfy the following equations:

$${x_1}^2 + {y_1}^2 = {x_2}^2 + {y_2}^2 = {x_3}^2 + {y_3}^2 \\ x_1 + x_2 + x_3 = 5 \\ y_1 + y_2 + y_3 = 0 \\ (x_1, y_1) \neq (x_2, y_2) \\ (x_1, y_1) \neq (x_3, y_3) \\ (x_2, y_2) \neq (x_3, y_3)$$ Currently, for each integer $$c$$ that can be written as the sum of two squares, I create a list of all possible integer pairs $$(a, b)$$ for which $$a^2 + b^2 = c$$ and then check all possible triplets from each list for validity (actually I only need to check all possible combinations of 2 pairs, but the idea is the same). The only constraints I've found are $$c$$ must be divisible by 5 and $$a$$ and $$b$$ must have different parity. This is pretty inefficient, as only a very small subset of possible values of $$c$$ produce valid triplets. Checking the validity of any given triplet is relatively easy, so I'm hoping someone can help me find further constraints on the values of $$c$$. It is also entirely possible that there is some other, more efficient method of finding these triplets. Either way, help would be much appreciated.

• If $c$ is a sum of two squares, then it's a square times a product of primes congruent to $1$ modulo $4$, or twice such a number. If it's a sum of relatively prime squares in three different ways, then it is divisible by at least three different primes congruent to $1$ modulo $4$. If $p=r^2+s^2$ and $q=t^2+u^2$, then $pq=(rt+su)^2+(ru-st)^2=(rt-su)^2+(ru+st)^2$. Nov 18, 2020 at 23:03
• Is your last chain of inequalities also supposed to mean that $(x_1,y_1)\neq(x_3,y_3)$? Nov 19, 2020 at 7:59
• Any thoughts on my comment, GOG? Nov 20, 2020 at 11:42
• $58$ total valid triplets? You have proved this? Or do you just mean so far you have found $58$ valid triplets, but there could be many more? Nov 20, 2020 at 22:52
• I get it now. There are $30$ useful values of $c$ (in that range) but some of them produce more than one valid triplet, so, $58$ valid triplets from those $30$ values of $c$. By the way, I can prove there are infinitely many valid triplets, in fact, infinitely many of the form $(p,0),(r,s),(r,-s)$, starting with $(5,0),(0,5)(0,-5)$; $(85,0),(-40,75),(-40,-75)$; $(1205,0),(-600,1045),(-600,-1045)$; $(16805,0),(-8400,14555),(-8400,-14555)$. Nov 21, 2020 at 5:11

We can use functions derived from the Pythagorean theorem where $$A^2+B^2=C^2.\quad$$ One generator is Euclid's formula, shown here as $$\quad A=m^2-k^2\quad B=2mk \quad C=m^2+k^2$$

All $$C$$-values are of the form $$4n+1$$ but not all $$4n+1$$ are valid $$C$$-values. A short list is shown here. You will notice, for $$C\in \{65,85,145,185,205,221,265,305,325,365\}$$, there are $$2$$ entries because those have $$2$$ different primitive Pythagorean triples each. For $$C=1105$$, there would be $$4$$.

We can find the $$(m,k)$$ pairs for triples for any given $$C$$-value by solving the $$C$$-function for $$k$$ and testing a defined range of $$m$$-values to see which yield integers. If none are found, there is no primitive or $$2\times$$ or square-multiple of a primitive for that $$C$$-value. Here is how we find triples.

$$\begin{equation} C=m^2+k^2\implies k=\sqrt{C-m^2}\qquad\\ \text{for}\qquad \bigg\lfloor\frac{ 1+\sqrt{2C-1}}{2}\bigg\rfloor \le m \le \lfloor\sqrt{C-1}\rfloor \end{equation}$$ The lower limit ensures $$m>k$$ and the upper limit ensures $$k\in\mathbb{N}$$.

$$C=65\implies \bigg\lfloor\frac{ 1+\sqrt{130-1}}{2}\bigg\rfloor=6 \le m \le \lfloor\sqrt{65-1}\rfloor=8\\ \quad\land \quad m\in\{7,8\}\Rightarrow k\in\{4,1\}\\$$ $$F(7,4)=(33,56,65)\qquad \qquad F(8,1)=(63,16,65)$$

The number of "primitive" triples will be $$\quad 2^{n-1}$$ where $$n$$ is the number of prime factors of $$C$$ meaning there will never be $$3$$ and only $$3$$ primitive triples for any $$C$$-value. For example, with $$C=65=5\times13$$, there are $$2^1=2$$ primitive triples. There $$are$$ values that also have non-primitive corresponding triples such as $$C=325$$. The first is non-primitive but those triples are

$$f(15,10)=(125,300,325)\quad f(17,6)=(253,204,325)\quad f(18,1)=(323,36,325)$$ So we have $$\qquad125^2+300^2\space =\space 253^2+204\space =\space 323^2+36^2\space =\space 325^2$$

The combination $$125+253-323\ne5\space$$ but $$\space 125+253-323=55$$ if that helps (kidding). Dividing both $$x$$ and $$y$$ by $$11$$ would be valid but $$300-204-36\ne0$$ so this triple of triple will not work in any case.

There are $$67$$ C-values where $$\quad C=4n+1\space\text{ for }\space 81\le n\le 11925\quad$$ that have $$3$$ triples each. You would have to find them programmatically as I did but perhaps one of them satisfies your requirement $$\quad x_2+x_2+x_3=5\quad\land\quad y_2+y_2+y_3=0$$