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xy + yz + zx = 3n^2 - 1, x + y + z = 3n , when x >=y>= z

Source: 1977 Swedish Mathematical Competition p3

March 26, 2021
diophantinesystem of equationsSystemnumber theory

Problem Statement

Show that the only integral solution to {xy+yz+zx=3n2−1x+y+z=3n\left\{ \begin{array}{l} xy + yz + zx = 3n^2 - 1\\ x + y + z = 3n \\ \end{array} \right. with x≥y≥zx \geq y \geq z is x=n+1x=n+1, y=ny=n, z=n−1z=n-1.
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