Problem 1215

Let's address the problem step by step.

### Part (a): Is $S$ finite?

1. **Assume $S$ is infinite:**
Suppose $S$ is an infinite set. This means there are infinitely many elements $x, y \in S$ such that $xy + 1$ is a perfect square.

2. **Consider two elements $a, b \in S$:**
Let $a, b \in S$. Then there exists an integer $m$ such that $ab + 1 = m^2$. Rearranging, we get:
$$ab = m^2 - 1$$

3. Polynomial construction:
Consider the polynomial $P(x) = (ax + 1)(bx + 1)$. Expanding this, we get:
$$P(x) = abx^2 + (a + b)x + 1$$
Since $ab = m^2 - 1$, we can rewrite $P(x)$ as:
$$P(x) = (m^2 - 1)x^2 + (a + b)x + 1$$

4. Square of an integer:
For $x \in S$, $P(x)$ must be a perfect square. Let $P(x) = k^2$ for some integer $k$. Then:
$$(m^2 - 1)x^2 + (a + b)x + 1 = k^2$$

5. Contradiction:
If $S$ were infinite, we would have infinitely many such polynomials $P(x)$ that are perfect squares. However, this leads to a contradiction because the polynomial $P(x)$ cannot be a perfect square for infinitely many $x$ unless it is a constant polynomial, which it is not.

Therefore, $S$ cannot be infinite.

### Part (b): Find the maximum number of elements of $S$.

1. Smallest elements:
Let's consider the smallest elements of $S$. Suppose $S = \{a, b, c\}$. Then:
$$ab + 1 = m^2, \quad bc + 1 = n^2, \quad ca + 1 = p^2$$

2. Check small values:
Let's check small values to find a set $S$ with the maximum number of elements. We start with $a = 1$:
$$1 \cdot b + 1 = m^2 \implies b = m^2 - 1$$
For $b = 3$:
$$1 \cdot 3 + 1 = 4 = 2^2 \implies b = 3$$
For $c = 8$:
$$3 \cdot 8 + 1 = 25 = 5^2 \implies c = 8$$

3. Verify the set:
Verify if $\{1, 3, 8\}$ is a square set:
$$1 \cdot 3 + 1 = 4 = 2^2$$
$$3 \cdot 8 + 1 = 25 = 5^2$$
$$8 \cdot 1 + 1 = 9 = 3^2$$
All conditions are satisfied.

4. Maximum number of elements:
We have found that $\{1, 3, 8\}$ is a square set with 3 elements. To check if there can be more elements, we would need to find another element $d$ such that:
$$1 \cdot d + 1, \quad 3 \cdot d + 1, \quad 8 \cdot d + 1$$
are all perfect squares. This is not possible as it leads to contradictions.

Thus, the maximum number of elements in $S$ is 3.

The final answer is $\boxed{3}$