# Is there an integer decomposing into four particular Pythagorean triplets?

I'm looking for a non-zero integer, say $$c$$ which is written as $$c^2=a_1^2+b_1^2=a_2^2+b_2^2=a_3^2+b_3^2=a_4^2+b_4^2$$ (the couples $$(a_i,b_i)$$ being ordered in ascending order, distinct and of strictly positive integers) and where couples verify two equalities.

Let us assume $$a_1b_1\leq a_2b_2\leq a_3b_3\leq a_4b_4$$.

The equalities are:

• $$a_1b_1=a_3b_3-a_2b_2$$,
• $$a_4b_4=a_3b_3+a_2b_2$$.

I looked at triplets where $$c$$ is factorized into odd primes that each factorize into two conjugated Gaussian integers but I get heavy calculations with no real sign of progress.

I also ran a computer program in case $$c^2$$ has exactly (to simplify the code) four decompositions in sum of two squares and up to $$2\times 10^6$$ there is no solution.

• Solutions can be written as follows. artofproblemsolving.com/community/… $p,s-$ Think of it as A Pythagorean triple. And then do the conversion..... maybe something will happen. Oct 31, 2020 at 4:55
• I checked for $c=5^3\times 13\times 17\times 29\times 37\times 41\times 53\times 61$, which decomposes in 7654 ways, and did not even find 3 triples $(a_1,b_1,c), (a_2,b_2,c), (a_3,b_3,c)$ with $a_1 b_1+a_2 b_2=a_3 b_3$ - never mind a fourth triple! Nov 5, 2020 at 22:53

All $$C$$-values are of the form $$4n-1.\quad$$ Here is a simple BASIC program that I used to generate the $$C$$-values (in my other answer) that have $$4$$ matching primitive triples each. Sometimes there will be more than $$4$$ but those extras will be non-primitive. There will always be $$2^{n-1}$$ primitive triples for a $$C$$-value where $$n$$ is the number or prime factors of $$C$$. For instance, $$1105=5\times13\times17$$ so there are $$2^2$$ primitives. The number of primitives will be $$2^{n-1}$$ where $$n$$ is the number of unique prime factors of $$C$$. After the "run", the next section shows which ones produce all primitives.

 100 print "input limit";
110 input l1
120 for n1 = 1 to l1
130     c1=4*n1+1
140     m1=int((1+sqrt(2*c1-1))/2)
150     m2=int(sqrt(c1-1))
160     c9=0
170     for m0=m1 to m2
180         k0=sqrt(c1-m0^2)
190         if k0=int(k0)
200            c9=c9+1
210         endif
220      next m0
230      if c9=4
240         print c1,
250      endif
310 next n1


Here is a sample run where $$l1=3000$$

1105 1625 1885 2125 2405 2465 2665 3145 3445 3485 3625 3965 4225 4505 4625 4745 5125 5185 5365 5785 5945 6205 6305 6409 6565 6625 7085 7225 7345 7565 7585 7625 7685 8177 8245 8585 8845 8905 9061 9125 9265 9605 9685 9805 9945 10205 10585 10865 10985 11125 11245 11285 11645 11713 11765

The following $$C$$-values have $$4$$ matching primitive Pythagorean triple each where

$$a_1^2+b_1^2=a_2^2+b_2^2=a_3^2+b_3^2=a_4^2+b_4^2=c^2$$

$$1105, 1885, 2405, 2465, 2665, 3145, 3445, 3485, 3965, 4505, 5185, 5365, 5785\\ 5945, 6205, 6305, 6409, 6565, 7085, 7345, 7565, 7585, 7685, 8177, 8245, 8585\\ 8845, 8905, 9061, 9565, 9605, 9685, 9805, 10205, 10585, 10865, 11245 11285\\ 11645, 11713, 11765, 12505, 12545,12665, 12805, 12905, 13345, 13481, 13505\\ 13949, 14065, 14645, 14705, 14885, 14965, 15145, 15385, 15457, 15665, 15805$$

We can find these triple by solving Euclid's formula for $$C$$ and testing a defined range of m-values to see which, if any, yield integers. $$\quad A=m^2-k^2,\quad B=2mk,\quad C=m^2+k^2\quad$$

Here is an example using $$C=64$$ to find two primitive triples. $$\qquad 1105$$ would have yielded $$4$$

$$$$C=m^2+k^2\implies k=\sqrt{C-m^2}\\ \text{for}\qquad \bigg\lfloor\frac{ 1+\sqrt{2C-1}}{2}\bigg\rfloor \le m \le \lfloor\sqrt{C-1}\rfloor$$$$ The lower limit ensures $$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\\ \land \quad m\in\{7,8\}\implies k\in\{4,1\}\\\\$$ $$F(7,4)=(33,56,65)\qquad \qquad F(8,1)=(63,16,65)$$

If we were to use $$C=1105$$, we would find

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

• Please see my comment to original question. The problem is homogeneous, so a solution for, say $c=1105$ will imply solutions for all multiples of $1105$. I tried very large products of many primes of form $4n+1$ (also much larger than the one I mention there) and didn't find solutions for even just 3 triples. I very much doubt there will be 4 triples with the required properties. Jan 5, 2021 at 17:55