# Can $x^2+y^2,y^2+z^2,z^2+x^2$ and $x^2+y^2+z^2$ all be square numbers?

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?

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

• 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? Jul 14, 2016 at 13:22