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Are there integers $a_1,a_2,a_3\in\Bbb N$ so that any two $a_i,a_j,i\not=j$ of them are the right side of a Phytagorean triple, i.e.

$$a_i^2+a_j^2=b_{ij}^2,\qquad\text{for some }b_{ij}\in\Bbb N\quad?$$

Idea: I tried to choose some highly composite number $c$ and define $a_3=2c$. Given two decompositions $c=c_ic_i',i=1,2$ with $c_i<c_i'$, I defined $a_i=c_i'^2-c_i^2$. In this way the pairs $(a_i,a_3),i=1,2$ will yield Pythagorean triples. But numeric computation suggests that $(a_1,a_2)$ will never do so.

Is there another approach or is it simply not possible? If there are such triples $a_i,i=1,2,3$, is there a procedure to generate infinitely many of them?

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    $\begingroup$ You can't have two odd because the sum of two odd squares is never a square. If all are even you can divide through by $2$, and get a smaller set. So if they exist, in the smallest set one is odd and two are even. $\endgroup$ – Mark Bennet Sep 29 '17 at 9:44
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    $\begingroup$ And similarly in the smallest set exactly two are multiples of $3$. $\endgroup$ – Especially Lime Sep 29 '17 at 9:50
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    $\begingroup$ en.wikipedia.org/wiki/Euler_brick $\endgroup$ – Oscar Lanzi Sep 29 '17 at 10:14
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    $\begingroup$ @OscarLanzi Thank you! This was exactly what I was looking for. $\endgroup$ – M. Winter Sep 29 '17 at 10:45

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