Hey Redhorse,
Conceptually, I think the best way to look at this is to note first that, when taking the square root of any product, you can split the product.
So...the square root of xy = (sqrt x) * (sqrt y)
Then bring in that, when taking the square root of a number, you're essentially breaking that number down into two identical components that would multiply together. So really you're taking the factors of that number and splitting apart pairs to allocate one to each "half" of the square root. So, for example:
sqrt 4 = sqrt 2*2 ---> one 2 goes to each "half" of the square root, so you have a clean 2 in each portion. Sqrt 4 = 2
But if you take the square root of 27, you have:
sqrt 27 = sqrt 3*3*3 ---> the first two 3s split apart evenly, but the last one doesn't have a "clean" pair, so you have to divide that one into sqrt 3. So you have 3(sqrt 3) in each "half" of the square root.
So...in order to have an integer as the square root of a number, you must have clean pairs of prime factors so that each "half" of the root gets its own prime factor. When you're left with an odd number of any given prime factor, then you have to take the root of that prime to split it evenly between each "half" of the root, and then you don't have an integer.
So for statement 1, knowing that sqrt (4x) is an integer means that:
sqrt (4x) = sqrt 2*2*x = integer
Well, the 2s split evenly, allocating one to each "half", leaving just that x. And for x to produce an integer, it needs to break into even pairs, so we can prove that the square root of x must be an integer.
You could also look at it this way: If sqrt (4x) is an integer, then sqrt 4 * sqrt x is an integer, and we know that sqrt 4 is 2, so 2(sqrt x) must be an integer. Sqrt x must be an integer!
For statement 2, the "pairs" logic may be the best way to look at it. When we see:
sqrt (3x) is not an integer
We already know that the 3 alone does not split into integer pairs, so that 3 is already enough to give us the noninteger portion. The x could or could not give us integer pairs, but either way that sqrt 3 that we have to allocate to both "halves" of the square root is going to provide that noninteger answer (unless x provides an odd number of 3s). So statement 2 is not sufficient...x could be a pair of factors or it could not be.
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Now, just to push the explanation a little further, let's say that statement 2 instead said that:
"Sqrt 3x = integer"
Here, we'd know that the 3 alone requires another 3 to give us a pair of prime factors that can split to form an integer square root. So x MUST, in this case, provide at least one 3. In addition, x would need to have pairs of prime factors to allocate evenly across both "halves" of the square root in order to create an integer root. So with this statement 2 it would be sufficient.
x could be 3:
3x = 3*3, and the square root splits the 3s
x could be 27, or 3*3*3:
3x = 3*(3*3*3), so we can split the 3s into pairs
x could not be 9:
3x = 3*(3*3) ---> we have an odd number of the same prime factor, so we can't split it into pairs.
Brian Galvin
GMAT Instructor
Chief Academic Officer
Veritas Prep
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