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.