XXVII Olimpiada Matemática Rioplatense

Let $$M(d)=\frac{(a_1+d)(a_2+d)\dots (a_n+d)}{a_1{a}_2\dots a_n}.$$ The key to the proof is to notice that the assumption that $M(d)$ is an integer for every non-negative integer $d$ implies indeed that $M(d)$ is an integer for all integer $d$: if $d<0$, consider $d^{\prime }=|d|a_1\dots a_n+d$; then, $d^{\prime }\ge |d|+d\ge 0$ and so, $M(d^{\prime })$ is integer. Then, $(a_1+d^{\prime })(a_2+d^{\prime })\dots (a_n+d^{\prime })\equiv 0(\mathrm{mod}{a}_1\dots a_n)$; since $d^{\prime }\equiv d(\mathrm{mod}{a}_1\dots a_n)$, we deduce that $(a_1+d)(a_2+d)\dots (a_n+d)\equiv 0(\mathrm{mod}{a}_1\dots a_n)$ and, therefore, $M(d)$ is an integer. We will now show that $a_k=k$ for every $1\le k\le n$. We proceed inductively. Assuming that $a_i=i$ for every $i<k$, for $k\ge 1$, we will show that $a_k=k$. Consider $M(-k)$. By the induction hypothesis, we have that $$M(-k)=\frac{(-1{)}^{k-1}(k-1)!(a_k-k)(a_{k+1}-k)\dots (a_n-k)}{(k-1)!a_k{a}_{k+1}\dots a_n}=\frac{(-1{)}^{k-1}(a_k-k)(a_{k+1}-k)\dots (a_n-k)}{a_k{a}_{k+1}\dots a_n}.$$ If $a_k>k$, then $0<a_j-k<a_j$ for every $k\le j\le n$, and so, we have that $$0<(a_k-k)(a_{k+1}-k)\dots (a_n-k)<a_k{a}_{k+1}\dots a_n,$$ which implies that $0<|M(-k)|<1$. This contradicts the fact that $M(-k)$ is an integer. Therefore, $a_k=k$, which completes the induction.

We conclude that $a_k=k$ for every $1\le k\le n$. To finish the proof, note that the condition in the statement holds for these values, since for every integer $d\ge 0$, we have $M(d)=\left(\genfrac{}{}{0ex}{}{n+d}{n}\right)$, which is integer.