China Southeastern Mathematical Olympiad

By $2x\mid A$, note that $A=4\cdot p_2^{p_2}\cdots p_n^{p_n}$. We may suppose that $x=2^{{\alpha }_1}{p}_2^{{\alpha }_2}\cdots p_n^{{\alpha }_n}$, where $0\le {\alpha }_1\le 1$, $0\le {\alpha }_i\le p_i$ ($i=2,3,\dots ,n$). Then, we have $$\frac{A}{x}=2^{2-{\alpha }_1}{p}_2^{p_2-{\alpha }_2}\cdots p_n^{p_n-{\alpha }_n}.$$ Hence, the number of different divisors of $\frac{A}{x}$ is $$(3-{\alpha }_1)(p_2-{\alpha }_2+1)\cdots (p_n-{\alpha }_n+1).$$ We know that $$(3-{\alpha }_1)(p_2-{\alpha }_2+1)\cdots (p_n-{\alpha }_n+1)=x=2^{{\alpha }_1}{p}_2^{{\alpha }_2}\cdots p_n^{{\alpha }_n}.\stackrel{◯}{1}$$ By induction on $n$, we shall prove that the array $({\alpha }_1,{\alpha }_2,\dots ,{\alpha }_n)$ satisfying $\stackrel{◯}{1}$ is $(1,1,\dots ,1)$ ($n\ge 2$).

(1) If $n=2$, then $\stackrel{◯}{1}$ becomes $(3-{\alpha }_1)(4-{\alpha }_2)=2^{{\alpha }_1}{3}^{{\alpha }_2}$, where ${\alpha }_1\in \{0,1\}$. If ${\alpha }_1=0$, then $3(4-{\alpha }_2)=3^{{\alpha }_2}$ which has no integer solution ${\alpha }_2$. If ${\alpha }_1=1$, then $2(4-{\alpha }_2)=2\cdot 3^{{\alpha }_2}$. We have ${\alpha }_2=1$. Thus, $({\alpha }_1,{\alpha }_2)=(1,1)$. That is, the conclusion is true for $n=2$.

(2) Suppose that the conclusion is true for $n=k-1$ ($k\ge 3$), then when $n=k$, $\stackrel{◯}{1}$ becomes $$\begin{gathered} (3-{\alpha }_1)(p_2-{\alpha }_2+1)\cdots (p_{k-1}-{\alpha }_{k-1}+1)(p_k-{\alpha }_k+1)\stackrel{◯}{2} \\ =2^{{\alpha }_1}{p}_2^{{\alpha }_2}\cdots p_{k-1}^{{\alpha }_{k-1}}{p}_k^{{\alpha }_k}. \end{gathered}$$ If ${\alpha }_k\ge 2$, considering $$0<p_k-{\alpha }_k+1<p_k,$$ $$0<p_i-{\alpha }_i+1\le p_i+1<p_k(1\le i\le k-1),$$ we see that the left-hand side of $\stackrel{◯}{2}$ cannot be divided by $p_k$, but the right-hand side of $\stackrel{◯}{2}$ is a multiple of $p_k$, which is a contradiction.

If ${\alpha }_k=0$, then $\stackrel{◯}{2}$ becomes $$\begin{gathered} (3-{\alpha }_1)(p_2-{\alpha }_2+1)\cdots (p_{k-1}-{\alpha }_{k-1}+1)(p_k+1)\stackrel{◯}{3} \\ =2^{{\alpha }_1}{p}_2^{{\alpha }_2}\cdots p_k^{{\alpha }_k}. \end{gathered}$$ Note that $p_2,p_3,\dots ,p_k$ are odd primes, thus, on the one hand, $p_k+1$ is even. So the left-hand side of $\stackrel{◯}{3}$ is even. On the other hand, the right side of $\stackrel{◯}{3}$ is odd. So ${\alpha }_1=1$. But then $3-{\alpha }_1=2$, so the left-hand side of $\stackrel{◯}{3}$ is a multiple of $4$, but the right-hand side of $\stackrel{◯}{3}$ is not, which is a contradiction.

By the above argument, we must have ${\alpha }_k=1$, and in $\stackrel{◯}{2}$, $$p_k-{\alpha }_k+1=p_k^{{\alpha }_k}=p_k.$$ Thus, $$(3-{\alpha }_1)(p_2-{\alpha }_2+1)\cdots (p_{k-1}-{\alpha }_{k-1}+1)=2^{{\alpha }_1}{p}_2^{{\alpha }_2}\cdots p_{k-1}^{{\alpha }_{k-1}}.$$ By the induction hypotheses, ${\alpha }_1={\alpha }_2=\cdots ={\alpha }_{k-1}=1$. Thus, ${\alpha }_1={\alpha }_2=\cdots ={\alpha }_{k-1}={\alpha }_k=1$, that is, the conclusion is true for $n=k$.

By (1) and (2), we conclude that $({\alpha }_1,{\alpha }_2,\dots ,{\alpha }_n)=(1,1,\dots ,1)$, so the integer required is $x=2p_2\cdots p_n=p_1{p}_2\cdots p_n$.