Let $R$ be a binary relation on the set $\{1, 2, \dots, 10\}$, where $(x, y) \in R$ if the product of $x$ and $y$ is square of an integer. Which of the following properties is/are satisfied by $R$?

A pair $(x, y) \in R \iff x \cdot y = k^2$ for some integer $k$.

1. Reflexive: For every $x \in \{1, 2, \dots, 10\}$, $x \cdot x = x^2$, which is the square of the integer $x$. Thus, $(x, x) \in R$ for all $x$, so $R$ is reflexive.

2. Symmetric: If $(x, y) \in R$, then $x \cdot y = k^2$. Since multiplication is commutative, $y \cdot x = x \cdot y = k^2$. Thus, $(y, x) \in R$, so $R$ is symmetric.

3. Transitive: Any positive integer $n$ can be uniquely factored as $n = s \cdot m^2$, where $s$ is square-free (the square-free part of $n$). Then $x \cdot y$ is a perfect square if and only if $x$ and $y$ have the same square-free part: $\text{sf}(x) = \text{sf}(y)$. Since the relation "having the same square-free part" is an equivalence relation, transitivity holds:

   $$\text{sf}(x) = \text{sf}(y) \text{ and } \text{sf}(y) = \text{sf}(z) \implies \text{sf}(x) = \text{sf}(z) \implies (x, z) \in R$$

   Hence, $R$ is transitive.

4. Antisymmetric: For antisymmetry, $(x, y) \in R$ and $(y, x) \in R$ must imply $x = y$. However, consider $x = 2$ and $y = 8$: $2 \times 8 = 16 = 4^2$, so $(2, 8) \in R$ and $(8, 2) \in R$, but $2

e 8$. Thus,$R$ is not antisymmetric.

Therefore, the properties satisfied by $R$ are (A) Reflexive, (B) Symmetric, and (C) Transitive.