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