Iranian Mathematical Olympiad

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 $$(Q(x){)}^2\le a^2(x+1)(x+2)\cdots (x+k)<(Q(x)+1{)}^2,\text{for large values of }x.$$ 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 $A$ ($F(\varnothing )=1$) 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 $A\Delta B\ne \varnothing ,S$ 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 $x$.