Question

求方程组①6x-y-z=20,②x^2+y^2+z^2=1979的所有正整数解

Answer 1

Y、Z地位是等价的.有解

X = 11,Y,Z = (3,43)

或

X = 13,Y,Z = (21,37)

从①易知 Y+Z = 4 + 6A ,X = 4 + A

设Y*Z = B,

B最小时,Y、Z = 1、3 + 6A,B = 3 + 6A

B最大时,Y = Z = (4 + 6A)/2,B = (2+3A)^2

代入②有

(4 + A)^2 + (4 + 6A )^2 -2B = 1979

结合B的范围,可得不等式,求得A的范围.

最终解得:

A = 7,B = 129,此时X = 11,Y,Z = (3,43)

A = 9,B = 777,此时X = 13,Y,Z = (21,37)