Problem 1100

If $k$ is even, using the lemma proved in the solution of the 7th problem of the Third Round, we can say that there exist an integer $a$ and some polynomial $Q(x) \in \mathbb{Z}[x]$ such that
$$(Q(x))^2 \le a^2(x+1)(x+2)\cdots(x+k) < (Q(x)+1)^2, \quad \text{for large values of } x.$$
Since $Q(x)$ is a perfect square for infinitely many values of $x$, we get
$$(Q(x))^2 = a^2(x+1)(x+2)\cdots(x+k).$$
This is a contradiction since $a^2(x+1)(x+2)\cdots(x+k)$ does not have any multiple roots.

For odd values of $k$, we claim that if $\prod_{i=1}^k (a+i)$ is a perfect square, there is some nonempty proper subset $A$ of $\{1, 2, \dots, k\}$ such that $\prod_{i \in A} (a+i)$ is a perfect square. To prove the claim, consider all of the subsets of $S = \{a+1, a+2, \dots, a+k\}$.

For any subset $A$ of $S$, let $F(A)$ be the product of elements of $A$ ($F(\emptyset) = 1$) and let $g(A)$ be the square free part of $F(A)$.

Note that all of the prime divisors of $g(\{a+1\}), g(\{a+2\}), \dots, g(\{a+k\})$ are less than $k$, because if some prime number $p \ge k$ divides two of the numbers $g(\{a+1\}), g(\{a+2\}), \dots, g(\{a+k\})$, for example $p|a+i$ and $p|a+j$, we obtain $p|j-i$ and hence $p < k$. If for some $i, 1 \le i \le k, p$ divides only $g(\{a+i\})$, this leads to a contradiction because $g(S) = 1$ is a perfect square, and the power of $p$ in the factorization of $\prod_{i=1}^k a + i$ should be even.

For any $A \subseteq S$, $g(A)|g(\{a+1\})g(\{a+2\})\cdots g(\{a+k\})$ and so the prime divisors of all $g(A)$'s are less than $k$. This implies that for each $A \subseteq S$, $g(A)$ is a divisor of $2 \times 3 \times \cdots \times p_{\pi(k-1)}$. So we have $2^{\pi(k-1)}$ cases for $g(A)$.

Since $k-1 > \pi(k-1)$, $2^k > 2 \times 2^{\pi(k-1)}$. This implies that there are two subsets $A, B \subseteq S$ such that $A \ne B, S-B$ and $g(A) = g(B)$ (note that $g(A) = g(S-A)$ because $g(S) = 1$). Now, $g(A) = g(B)$, therefore $F(A)F(B)$ is a perfect square. We have
$$F(A)F(B) = F(A \cap B)^2 F(A\Delta B).$$
Hence $F(A\Delta B)$ is a perfect square. But $A\Delta B \ne \emptyset, S$ since $A \ne B, S-B$, and this is the desired subset.

By replacing $A\Delta B$ with $S - A\Delta B$ if necessary, we will get a nonempty proper subset $X$ of $\{1, 2, \dots, k\}$ with an even number of elements such that $\prod_{i \in X} (x + i)$ is a perfect square. Using the lemma in the solution of the 7th problem of the Third Round again, this product can be a perfect square for at most a finite number of values of $x$. Thus if $k$ is odd, then $(x+1)(x+2)\cdots(x+k)$ is a perfect square for at most a finite number of values of $x$.