Closest Divisors

For each candidate x in {num + 1, num + 2}, the closest factor pair is found by searching downward from floor(sqrt(x)); the first divisor d found gives the complementary factor x / d and minimizes their gap for that x. Compare the two resulting gaps and return the better pair. Searching from one rather than from the square root is correct but wasteful. Explain the square-root fact: factor pairs move farther apart as the smaller factor moves below sqrt(x). Check perfect squares and the small values of num. There is no need to factor num itself because the contract asks for factors of num + 1 or num + 2. The time bound is O(sqrt(num)) in the worst case and O(1) extra space.