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user44191
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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_3^2+b_3^2$$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.

I'm looking for a non-zero integer, say $c$ which is written as $c^2=a_1^2+b_1^2=a_2^2+b_2^2=a_3^2+b_3^2=a_3^2+b_3^2$ (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.

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.

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user155516
user155516

I'm looking for a non-zero integer, say $c$ which is written as $a_1^2+b_1^2=a_2^2+b_2^2=a_3^2+b_3^2=a_3^2+b_3^2$$c^2=a_1^2+b_1^2=a_2^2+b_2^2=a_3^2+b_3^2=a_3^2+b_3^2$ (the couples $(a_i,b_i)$ being ordered in ascending order, distinct and of strictly positive integers) and where couples checkverify 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 $10^6$$2\times 10^6$ there is no solution.

I'm looking for a non-zero integer, say $c$ which is written as $a_1^2+b_1^2=a_2^2+b_2^2=a_3^2+b_3^2=a_3^2+b_3^2$ (the couples $(a_i,b_i)$ being ordered in ascending order, distinct and of strictly positive integers) and where couples check 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 $10^6$ there is no solution.

I'm looking for a non-zero integer, say $c$ which is written as $c^2=a_1^2+b_1^2=a_2^2+b_2^2=a_3^2+b_3^2=a_3^2+b_3^2$ (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.

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user155516
user155516

Is there an integer decomposing into four particular Pythagorean triplets?

I'm looking for a non-zero integer, say $c$ which is written as $a_1^2+b_1^2=a_2^2+b_2^2=a_3^2+b_3^2=a_3^2+b_3^2$ (the couples $(a_i,b_i)$ being ordered in ascending order, distinct and of strictly positive integers) and where couples check 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 $10^6$ there is no solution.