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Re: The integer a is greater than 1 and is not equal to the squ
[#permalink]
18 Apr 2024, 03:08

For instance consider a=13, which follows the required condition

so, 2a=26 which is not a square of any integer

So, this discards the last option.

Pls explain....

so, 2a=26 which is not a square of any integer

So, this discards the last option.

Pls explain....

Re: The integer a is greater than 1 and is not equal to the squ
[#permalink]
18 Apr 2024, 03:38

Expert Reply

Since a is not the square of an integer, its square root cannot possibly be the square of an integer (in fact, the square root of a cannot even be an integer itself, since a is not a perfect square). This rules out choice (A).

Choices (B) and (C) can be eliminated by the following logic: if a is greater than 1, a^2 will be at least 4. Additionally, since a is an integer, a^2 will be a perfect square. There are no different perfect squares that are one unit apart (other than 0 and 1), so a^2 + 1 and a^2 – 1 cannot possibly be perfect squares (they are each one unit away from a perfect square that cannot be 0 or 1).

Choice (D) can be eliminated for a similar reason. If a = 2, then a^2 = 4, and the nearest lower perfect square is 1, 3 units away (3 is greater than 2). If a = 3, then a^2 = 9, and the nearest lower perfect square is 4, 5 units away (5 is greater than 3). If a = 4, then a^2= 16, and the nearest lower perfect square is 9, 7 units away (7 is greater than 4). This pattern shows that subtracting a from a^2 will result in a number somewhere in between a^2, a perfect square, and the next lower perfect square which is (a – 1)^2.

Choice (E) must be a perfect square, because the expression can be factored as (a – 1)^2. Since a is an integer, a – 1 is an integer and (a – 1)^2 is a perfect square.

Finally, choice (F) can be a perfect square whenever a is equal to half of a perfect square. For example, if a = 2 or 8, then 2a = 4 or 16, respectively—both of which are perfect squares.

Choices (B) and (C) can be eliminated by the following logic: if a is greater than 1, a^2 will be at least 4. Additionally, since a is an integer, a^2 will be a perfect square. There are no different perfect squares that are one unit apart (other than 0 and 1), so a^2 + 1 and a^2 – 1 cannot possibly be perfect squares (they are each one unit away from a perfect square that cannot be 0 or 1).

Choice (D) can be eliminated for a similar reason. If a = 2, then a^2 = 4, and the nearest lower perfect square is 1, 3 units away (3 is greater than 2). If a = 3, then a^2 = 9, and the nearest lower perfect square is 4, 5 units away (5 is greater than 3). If a = 4, then a^2= 16, and the nearest lower perfect square is 9, 7 units away (7 is greater than 4). This pattern shows that subtracting a from a^2 will result in a number somewhere in between a^2, a perfect square, and the next lower perfect square which is (a – 1)^2.

Choice (E) must be a perfect square, because the expression can be factored as (a – 1)^2. Since a is an integer, a – 1 is an integer and (a – 1)^2 is a perfect square.

Finally, choice (F) can be a perfect square whenever a is equal to half of a perfect square. For example, if a = 2 or 8, then 2a = 4 or 16, respectively—both of which are perfect squares.

gmatclubot

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