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I know that if we want $x^2+y^2$ to be square number, we are looking for pythagorean triple; if we want $x^2+y^2+z^2$ to be a square number, we are looking for pythagorean quadruple. But have we ever found any positive integers $x,y,z$ such that $x^2+y^2,y^2+z^2,z^2+x^2,x^2+y^2+z^2$ are all square numbers?

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  • $\begingroup$ Might be nice to see an example of a Pythagorean quadruplet, as this is not something commonly known (in opposed to Pythagorean triplet). $\endgroup$
    – barak manos
    Jul 14, 2016 at 13:11
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    $\begingroup$ @barakmanos $3^2+4^2+12^2=13^2$ $\endgroup$
    – Hagen von Eitzen
    Jul 14, 2016 at 13:11
  • $\begingroup$ What is the source of this problem?(if there is any) $\endgroup$
    – Roby5
    Jul 14, 2016 at 13:14
  • $\begingroup$ For numbers you can write. artofproblemsolving.com/community/… $\endgroup$
    – individ
    Jul 14, 2016 at 13:16
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    $\begingroup$ @HagenvonEitzen: $1^2+2^2+2^2=3^2$. $\endgroup$
    – user21820
    Jul 14, 2016 at 13:24

1 Answer 1

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It's a famous open problem : the perfect cuboid problem.

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  • $\begingroup$ That's another question. $\endgroup$
    – individ
    Jul 14, 2016 at 13:15
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    $\begingroup$ I don't follow: a perfect cuboid is a cuboid whose edges $x$, $y$ and $z$ are integers and such that the edge diagonals and the main diagonal are all integers, which is equivalent to $x^2 + y^2$, $x^2 + z^2$ and $y^2 + z^2$ on the one hand, and $x^2 + y^2 + z^2$ all being perfect squares. Isn't that exactly the OP's question? $\endgroup$
    – PseudoNeo
    Jul 14, 2016 at 13:22
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    $\begingroup$ @PseudoNeo Oops. Sorry my mistake. I thought space diagonals didn't include edge diagonals. $\endgroup$
    – Roby5
    Jul 14, 2016 at 13:26

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