Skip to main content
MathOverflow

Return to Answer

edited body
Source Link
John R Ramsden
John R Ramsden
  • 1.5k
  • 13
  • 20

The equation is equivalent to $(2 x + 1)^2 + 1 = (2 y + 1)^2 + (2 z + 1)^2$. So given that an integer solution to this is implies integers $a, b, c, d$ with exactly one odd and $(a, d) = (b, c) = 1$ with $2 x + 1, 2 y + 1, 2 z + 1 = a c + b d, a b + c d, a c + c d$$2 x + 1, 2 y + 1, 2 z + 1 = a c + b d, a b + c d, a c - b d$ subject to $a b - c d = 1$, it would appear the original is equivalent to the latter.

The equation is equivalent to $(2 x + 1)^2 + 1 = (2 y + 1)^2 + (2 z + 1)^2$. So given that an integer solution to this is implies integers $a, b, c, d$ with exactly one odd and $(a, d) = (b, c) = 1$ with $2 x + 1, 2 y + 1, 2 z + 1 = a c + b d, a b + c d, a c + c d$ subject to $a b - c d = 1$, it would appear the original is equivalent to the latter.

The equation is equivalent to $(2 x + 1)^2 + 1 = (2 y + 1)^2 + (2 z + 1)^2$. So given that an integer solution to this is implies integers $a, b, c, d$ with exactly one odd and $(a, d) = (b, c) = 1$ with $2 x + 1, 2 y + 1, 2 z + 1 = a c + b d, a b + c d, a c - b d$ subject to $a b - c d = 1$, it would appear the original is equivalent to the latter.

Source Link
John R Ramsden
John R Ramsden
  • 1.5k
  • 13
  • 20

The equation is equivalent to $(2 x + 1)^2 + 1 = (2 y + 1)^2 + (2 z + 1)^2$. So given that an integer solution to this is implies integers $a, b, c, d$ with exactly one odd and $(a, d) = (b, c) = 1$ with $2 x + 1, 2 y + 1, 2 z + 1 = a c + b d, a b + c d, a c + c d$ subject to $a b - c d = 1$, it would appear the original is equivalent to the latter.