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If you take a quaternion number lets call it Q (Q=a+bi+cj)) Q² definitely generate a pythagorean quadruple and Q=a+bi+cj+dk will definitely generate pythagorean quintuple

ex: Q=1+1i+1j+k, Q² = (-2)+2i+2j+2k, and 2²+2²+2²+2²=16 and 16 is 4² and this is true for any integers a,b,c,d

proof 1: Q = a+bi+cj+dk, Q² = (a+bi+cj+dk)(a+bi+cj+dk)= (a² - b² - c² - d²) + 2abi + 2acj + 2adk remove all the complex numbers so it'll be (a² - b² - c² - d²) + 2ab + 2ac + 2ad square every part so it'll be (a² - b² - c² - d²)² + (2ab)² + (2ac)² + (2ad)² and if you factor it it'll be (a²+b²+c²+d²) and there is it 4 squared numbers that equal another square

Itay the king
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  • Please define pythagorean $n$-tuple, and state precisely the result you mean. – Kimball Oct 10 '22 at 10:47
  • Yes, this is all well-known. @Kimball I think it's pretty clear what the OP means, though their English is poor. – anon Oct 11 '22 at 11:17
  • @runway44 It's not clear to me--e.g., if $Q$ satisfies $d=0$ in their notation, $Q^2$ typically does not, so I'm not sure whether what they want to say in this case is true. – Kimball Oct 11 '22 at 14:01

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