Awesome question.

First we need to establish the numbers we can plug in for $a$.

If a number $x$ is not a square of an integer, then $x ≠ y^2$, where $y$ is any integer. So plug in integers for $y$, and they can't be $a$.

4,9,16,25,36..... can't be used.

The question is, if we plug in any other integers for $a$, will they result in a square of an integer? (4,9,16,25,36....)


❑ $\sqrt{a}$

This is incorrect, since the only way this yields an integer is if $a$ is a square of an integer, and we know it can't be.
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❑ $a^2$ $- 1$

This is incorrect because no square of an integer exists that is one integer away on the number line from it, other than 1.
Think $2^2 - 1 = 3$, or $3^2 - 1 = 8$. We have our square already, and we're subtracting one away from it, so the result can't be a square.

If $a = 1$, then it could be, since:

$1^2 - 1 = 0$

And 0 is a square of an integer ( $\sqrt{0} = 0$ ). But we know that $a ≠ 1$, so this choice must be incorrect.
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❑ $a^2$ $+ 1$

This is incorrect for the same reason that the above is incorrect. This time, not even $a = 1$ helps this one.
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❑ $a^2$ $- a$

Again, using similar logic as above, this is incorrect. However 1 would work for this case:

$1^2 - 1 = 0$
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❑ $a^2$ $- 2a + 1$

This one is correct. To see it, you have to factor it:

$(a-1)^2$

Now we can let $a = 5$, and we get 16, and 16 is a square of an integer.
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❑ $2a$

This one is also correct. Plug in 8 to get:

$2(8) = 16$, and 16 is a square of an integer.
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So the answers are E and F