This is solvable if and only if x has a perfect square factor.
proof:
x = a + b + sqrt(ab) so ab must be square
if a has an odd powered factor then b has the same odd powered factor.
sqrt(a) simplifies to a number of form d*sqrt(c) implied sqrt(b) simplifies to e*sqrt(c)
therefore
sqrt(x) = d*sqrt(c) + e*sqrt(c) = (d+e)*sqrt(c)
x = (d+e)^2 * c
so x has a perfect square factor.
Counting the number of integers which have a square factor takes awhile but is trivial.
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