2015 Thirteenth IMAR Mathematical Competition

The integers greater than $3$ are ruled out by noticing that if $a_1,a_2,a_3$ are pairwise distinct positive integers, then $$\begin{array}{rl}\frac{(a_1+1)(a_2+1)(a_3+1)-1}{a_1{a}_2{a}_3}& =1+\frac{1}{a_1}+\frac{1}{a_2}+\frac{1}{a_3}+\frac{1}{a_1{a}_2}+\frac{1}{a_1{a}_3}+\frac{1}{a_2{a}_3}\\ & \le 1+1+\frac{1}{2}+\frac{1}{3}+\frac{1}{1\cdot 2}+\frac{1}{1\cdot 3}+\frac{1}{2\cdot 3}=3+\frac{5}{6}\end{array}$$ To complete the proof, fix an integer $n\ge 3$, consider an integer $a\ge 3$, and let $a_1=a-2$, let $a_k=a^{2^{k-2}}$, $k=2,\dots ,n-1$, and let $a_n=a^{2^{n-2}}-2$. Then $1\le a_1<a_2<\cdots <a_n$, and $$\begin{array}{rl}\frac{(a_1+1)(a_2+1)\cdots (a_n+1)-1}{a_1{a}_2\cdots a_n}& =\frac{((a-1){\prod }_{k=2}^{n-1}(a^{2^{k-2}}+1))(a^{2^{n-2}}-1)-1}{(a-2)a^{1+2+\cdots +2^{n-3}}(a^{2^{n-2}}-2)}\\ & =\frac{{\left(a^{2^{n-2}}-1\right)}^2-1}{(a-2)a^{2^{n-2}-1}(a^{2^{n-2}}-2)}=\frac{a^{2^{n-2}}(a^{2^{n-2}}-2)}{(a-2)a^{2^{n-2}-1}(a^{2^{n-2}}-2)}\\ & =\frac{a}{a-2}=1+\frac{2}{a-2},\end{array}$$ which is integral if and only if $a=3$ or $a=4$. The former shows that $3$ satisfies the required condition, and the latter shows that so does $2$.