Are there any 2 primitive pythagorean triples who share a common leg? So is it possible for:
$\gcd(a,b,c)=1$
$a^2+b^2=c^2$
and
$\gcd(a,d,e)=1$
$a^2+d^2=e^2$
?

This system of equations:
$$\left\{\begin{aligned}&a^2+b^2=c^2\\&z^2+b^2=x^2\end{aligned}\right.$$
Solutions have the form:
$$b=4tkp^2s^2$$
$$a=2(t^2-k^2)p^2s^2$$
$$z=4k^2s^4-t^2p^4$$
$$c=2(t^2+k^2)p^2s^2$$
$$x=4k^2s^4+t^2p^4$$
$t,k,p,s$ - integers asked us.
