Let integer v > 1 be a perfect square, so v = k^2 for some integer k geq 2, and sqrt v = k. Which of the following expressions must also be a perfect square? Select all that apply.

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GRE Quantitative Reasoning QUESTION #8194

Question 1 (Select all that apply)

Let integer $v > 1$ be a perfect square, so $v = k^2$ for some integer $k \geq 2$, and $\sqrt{v} = k$. Which of the following expressions must also be a perfect square? Select all that apply.

$81v$✔️
$25v + 10\sqrt{v} + 1$✔️
$4v^2 + 4\sqrt{v} + 1$
$v + 2\sqrt{v} + 1$✔️

Correct Answer Explanation

Since $v = k^2$ and $\sqrt{v} = k$:

Option A — $81v$: $= 81k^2 = (9k)^2$. Always a perfect square. ✓

Option B — $25v + 10\sqrt{v} + 1$: $= 25k^2 + 10k + 1 = (5k+1)^2$. Always a perfect square. ✓

Option C — $4v^2 + 4\sqrt{v} + 1$: $= 4k^4 + 4k + 1$. Test $k = 2$: $4(16) + 4(2) + 1 = 73$. $\sqrt{73}$ is not an integer. Not always a perfect square. ✗

Option D — $v + 2\sqrt{v} + 1$: $= k^2 + 2k + 1 = (k+1)^2$. Always a perfect square. ✓

Correct answers: A, B, and D (indices 0, 1, 3).

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