If x is a positive integer, which of the following CANNOT be expressed

If we take x = 2 & calculate, we cant get the answers;
seems that x has to be taken 1 to execute all the options

So I guess zero does count as a perfect square then

Cheers
J

Even though Bunuel and karishma have answered this question, I am not able to digest any fundamental of this ....

Not sure how to crack this one... as I got this question in my Veritas prep exam today...

Look, the question simply asks which option CANNOT be a perfect square. In the options, x is a positive integer.

Can x^5 be a perfect square? Can x take some value such that x^5 is a perfect square? Say, x = 4. Then x^5 = 4^5 = 2^10
This is a perfect square. Hence for some value of x, x^5 could be a perfect square. Hence this is not our answer.
How do we find a value for which x^5 will be a perfect square? Perfect squares have even powers. We have x^5 which is an odd power. To get an even power, we could select x such that it already has an even power - we selected x = 2^2. Similarly, x could be 1^2 or 3^2 or 4^2 or 5^2 or 3^4 etc

Now think, can x^2 - 1 be a perfect square?
The reasoning for all the options is given in the post above.

A more intuitive approach is putting x = 1 as given by Bunuel. When x = 1, all options except option (D) results in a perfect square. So we know that all options CAN be perfect squares except (D). By elimination, answer must be (D).

So the one point where I get tripped up is the X^2 - 1. 0 is assumed to be a perfect square?

How can I tell from the verbage of the question that what they are asking for is determining whether or not something is a perfect square or not?

Thanks for the help.

Yes, both 0 and 1 are perfect squares.

"Can you express 'this' as n^2 where n is an integer?" asks us whether we can write 'this' as square of an integer.
Square of an integer is a perfect square. So the question becomes "Can you express 'this' as a perfect square?"

Slightly convoluted verbiage is common in GMAT.

Please refer to option [b] where value comes 0.
Would you please clarify whether 0 is a perfect square?

Yes, 0 is a perfect square 0 = 0^2.

I picked B, because I thought that 2 consecutive integers multiplied do not equal a perfect square...I did not take into consideration the possibility of 0...

I think the hardest part about this question is working out what is being asked.

Which of the following cannot be expressed as n^2 means which of the following cannot be perfect square when n is an integer.

Subbing in 1 for x in each example brings us to answer D.

1^2 + 1 = 2, which is not the square of any integer. D is your answer! Easy

D.

x^2+1 = n^2 => x^2 = (n+1)(n-1) and n>0. This can never give a perfect square

Official explanation:

D. For each of choices A, B, C, and E, trying the value x = 1 shows that there is one possible value of x for which the expression results in an integer squared:

A) The result is 1, which is the same thing as 1^2
B) The result is 0, which is the same thing as 0^2
C) The result is 1, which is the same thing as 1^2
E) The results is 1, which is the same thing as 1^2
For D, however, you cannot find a value of x for which x2+1
will yield a perfect square. If x = 1, the expression yields a 2. If x = 2, the expression yields a 5. As no two squares (other than 0 and 1) are one number apart, D cannot be a perfect square if x is a positive integer.

A. If \(x = 4\), \(x^5\) can be expressed as a perfect square.

B. \(x^2 - 1 = n^2\)
x must be equal to 1.

C. \(x^4 = n^2\)
\(x^2 = n\)

D. \(x^2 + 1 \)
If \(x = 1\), then \(x^2 + 1 = 2\). Not a perfect square.

E. \(\sqrt{x^5}\) = \(\sqrt{1^5}\) = 1

It's not necessary that x be 1, that's just the easiest option.

In answer choice a, we could let \(x=n^\frac{2}{5}\), for example.

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