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標題:

Number Theory

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發問:

Find all integers (x,y) satisfying x^2(z-y) +y^2(x-z) +z^2(y-x) = 1125

最佳解答:

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.

其他解答:

(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