If $$x, y, z$$ are positive integers satisfying the system of equations $$xy + yz + zx = 2024$$ and $$xyz + x + y + z = 2025$$, find $$\max(x, y, z)$$.

Subtracting the first equation from the second gives $$xyz - (xy+yz+zx) + (x+y+z) - 1 = 0$$, which is exactly $$(x-1)(y-1)(z-1) = 0$$, so one variable equals 1. Taking $$x = 1$$ reduces the first equation to $$y + z + yz = 2024$$, that is $$(1+y)(1+z) = 2025$$. The factorisation $$2025 = 3 \times 675$$ gives $$y = 2$$ and $$z = 674$$, the largest value the maximum can take.