solve x^2+Y^=81

Question

I am having difficulty finding out why the anwer to this question is zero

Answer 1

y²=81-x²=(9-x)(9+x) where 0≤x≤9 because a perfect square (y² in this case) must be positive.

If y is an integer we just substitute all possible integer values for x and see if we get the perfect square of an integer:

x=9, y=0; x=8, y=√10; x=7, y=4√2; x=6, y=√45; x=5, y=2√14; x=4, y=√65; x=3, y=6√2; x=2, y=√77; x=1, y=4√5. The only integer value for y is zero (when x=9, or vice versa). But the question doesn't say that x and y have to be integers, so there are many possible solutions y=√(81-x²) for 0≤x≤9.