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