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Number with a sixth root will have an integer square root

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Source: — Data Sufficiency |
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by MartyMurray » Fri Jan 13, 2017 12:18 am
Notice that the question is not "will the answer always be an integer?"

The question is "could the answer be an integer?"

Any number with an integer sixth root will also have an integer square root, as the square root is the cube of the sixth root.

So for any number such that the sixth root is an integer, x¹/� + x¹/² will be an integer.

Another way to look at is that for any number x such that x = an integer raised to the sixth power, x¹/� + x¹/² will be an integer.
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by Brent@GMATPrepNow » Mon Jan 16, 2017 9:17 am
bartwalhari wrote:For each of the following, could the answer be an integer if x is an integer greater than 1?
a) x^10 + x^(-10) =
b) x^(1/6) + x^(1/2) =
Odd wording for a DS question! It would never appear on the GMAT.

That said, I just wanted to point out that, if x = 64, then x^(1/6) + x^(1/2) = 64^(1/6) + 64^(1/2) = 2 + 8 = 10

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by Matt@VeritasPrep » Fri Jan 20, 2017 12:28 am
Another thing worth pointing out here is that

(x¹/�)³ = (x¹/²)

So if x¹/� is an integer, x¹/² will be as well.
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