Track / Stage 3 / 80 of 260 #80 of 1964

Problem 80

AMC 10/12, early questions

Number theory Difficulty 3.3 Multiple choice

Let the set consisting of the squares of the positive integers be called $u$ ; thus $u$ is the set $1, 4, 9, 16 \ldots$ .
If a certain operation on one or more members of the set always yields a member of the set,
we say that the set is closed under that operation. Then $u$ is closed under:

Pick one

Official solution

Consider each option, case by case.
For option A, note that $1 + 4 = 5$ , so the set is not closed under addition because $5$ is not a perfect square.
For option B, note that $9 \cdot 4 = 36$ and $4 \cdot 25 = 100$ . Letting one member of the set be $a^2$ and another member be $b^2$ (where $a$ and $b$ are positive integers), the product of the two members is $(ab)^2$ . Since $ab$ is an integer and $(ab)^2$ is a perfect square, the set is closed under multiplication.
For option C, note that $9 \div 4 = 2.25$ , so the set is not closed under division because $2.25$ is not an integer.
For option D, note that $\sqrt{4} = 2$ , so the set is not closed under extraction of a positive integral root because $2$ is not a perfect square.
Thus, the answer is $\boxed{\textbf{(B)}}$ .

Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic, difficulty and ordering added by this site.