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?