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Good question (and congratulations on the lofty Fuqua ambitions!). In this case, asking whether n is equal to the square of an integer is essentially asking whether n is the result of squaring another integer. For example:
16 is the square of an integer (4^2 = 16).
2 is not the square of an integer (the square root of 2 is 1.4......, and not an integer)
Another way to look at it is to ask whether the square root of the n is an integer - that's really the same question. If n is the result of an integer squared, then the square root of n would be that integer.
Knowing that, you might use the "Jump Ahead" strategy from the Data Sufficiency lesson to assess Statement 2 first, as Statement 2 provides precisely that information - by saying that the square root of n is an integer, you know definitively that n is the result of an integer squared.
With Statement 1, the notion that "for any prime factor of n, that factor squared is also a factor" is very close to being sufficient. If every prime factor were also a factor when squared, you'd be close to the form that n = (xy)^2, where the factors are x, y, x^2, and y^2. But, this isn't entirely perfect, as you can take a number like 32 and break it down in to prime factors: 2*2*2*2*2 (2^5). 2 is its only prime factor, and 2^2, or 4, is also a factor. But, 32 is not the square of an integer. The catch is that p^2 could be a factor of p^3 -- for n to be the square of an integer, you'd need an even exponent for p, as the square root of an odd exponent won't be an integer.
That's a pretty technical explanation - the easy way to look at is to try and prove insufficiency by looking at two numbers that would both satisfy the statement. 4 does, and gives us the answer "yes" (n = 4, p is 2, and p^2 is also a factor; 4 is the square of an integer). 8 does, as well, but gives us the answer "no" (n = 4, p = 2, and p^2 is also a factor, but 8 is not the square of an integer).
Because statement 1 is not sufficient, but statement 2 is, the answer is B.
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