Do there exist $3$ different primitive Pythagorean triples $(a,d,w), (a,b,z)$ and $(c,d,z)$?

Explicitly, we want $6$ different integers $a,b,c,d,w,z$ such that...

(1) $a^2 + d^2 = w^2$

(2) $a^2 + b^2 = c^2 + d^2 = z^2$

  • $\begingroup$ Apologies for the confusion caused by my initial post (I really did try to not make a mistake, sigh), I've now edited it. $\endgroup$ – fuzzy Jan 26 '18 at 11:05
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    $\begingroup$ I have written a piece of code that is currently looking, and it hasn't found any examples so far with $z\leq 350\,000$. Assuming the code is correct... It's still running, but it takes longer and longer to check. $\endgroup$ – Arthur Jan 26 '18 at 12:38

Unless I've made a mistake with my arithmetic these 6 will do...

a = 7015 b = 46248 c = 46535 d = 4752 w = 8473 z = 46777


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