0≤√x≤√336、0≤√y≤√336，

Rewrite the original form as follows

y=(4√2 1-√

x)^2

=

336+

x

-

8√2 1 √

x

Y is an integer, so we can know that X is the product of 2 1 and the square of another integer, and we can set X = 2 1 A 2.

(a is an integer)

∴

y=2 1(4-a)^2

According to the range of x and y, we can know that 0 ≤ 2 1A 2 ≤ 336, 0 ≤ 2 1 (4-A) 2 ≤ 336.

Solution: 0≤a≤4

∴

X and y have five sets of values.