P and q are odd numbers.
And x 1 is an integer.
x 1+x2=-P
∴x2 and x 1 are even and odd numbers.
x 1*x2=q
∴x2 and x 1 are odd numbers.
The two conditions are contradictory, so the three-dimensional equation cannot have an integer solution.
2. Assume that (1/x2-1) (1/y2-1) ≥ 9 holds.
(1/x2-1) (1/y2-1) ≥ 9 can be changed to
Can it be converted into 1/x? y? -( 1/x? + 1/y? )+ 1≥9
( 1-(x? +y? ))/x? y? ≥8
( 1-(x+y)? +2xy)/x? y? ≥8
2/xy≥8
xy≤ 1/4
∵X, Y>0 and x+y= 1.
∫x+y≥2√xy = 1
That is 2√xy≤ 1.
4xy≤ 1
xy≤ 1/4
The hypothesis that (1/x2-1) (1/y2-1) ≥ 9 holds.