Number Theory

Let f(x, y, z) = x^2(z - y) + y^2(x - z) + z^2(y - x)
since f(y, y, z) = y^2(z - y) + y^2(y - z) + z^2(y - y) = 0
x - y is a factor of f(x, y, z)
without loss of generality, y - z and z - x are factors of f(x, y, z)
thus, f(x, y, z) = k(x - y)(y - z)(z - x)
by comparing the coefficients, we have k = 1
So, x^2(z - y) + y^2(x - z) + z^2(y - x) = (x - y)(y - z)(z - x)

x^2(z - y) + y^2(x - z) + z^2(y - x) = 1125
(x - y)(y - z)(z - x) = 1125
for x is odd, y is odd, z is odd, (x - y)(y - z)(z - x) is even
for x is odd, y is odd, z is even, (x - y)(y - z)(z - x) is even
for x is odd, y is even, z is odd, (x - y)(y - z)(z - x) is even
for x is odd, y is even, z is even, (x - y)(y - z)(z - x) is even
for x is even, y is odd, z is odd, (x - y)(y - z)(z - x) is even
for x is even, y is odd, z is even, (x - y)(y - z)(z - x) is even
for x is even, y is even, z is odd, (x - y)(y - z)(z - x) is even
for x is even, y is even, z is even, (x - y)(y - z)(z - x) is even

(x - y)(y - z)(z - x) cannot be equal to a odd number
so there are no solutions.

(x－y)(y－z)(z－x)=1125
設x>=y>=z
x－y>=0，y－z>0，z－x<=0
(x－y)(y－z)(z－x)<=0與(x－y)(y－z)(z－x)=1125矛盾