number theory - Prove that there are positive integers $x_1,x_2,ldots,x_n geq 1$ such that $x_1x_2ldots x_n = a_1x_1+a_2x_2+cdots+a_nx_n$

Prove that for any natural number $n$ and for any natural numbers $a_k, k = 1,\ldots,n,\{a_1,a_2,\ldots,a_n\} = \{1,2,\ldots,n\},$ there are positive integers $x_1,x_2,\ldots,x_n \geq 1$ such that $$x_1x_2\ldots x_n = a_1x_1+a_2x_2+\cdots+a_nx_n.$$

Since $\{a_1,a_2,\ldots,a_n\} = \{1,2,\ldots,n\}$, we may assume that $a_k = k$ and our equation becomes $$x_1+2x_2+3x_3+\cdots+nx_n = x_1x_2 \cdots x_n.$$ For example, if $n = 2$ we need $x_1+2x_2 = x_1x_2$. Then we have $x_1(x_2-1) = 2x_2$ and so $x_1 = \dfrac{2x_2}{x_2-1}$, so $x_2 = 2$ and $x_1 = 4$. For $n = 3$ we need $$x_1+2x_2+3x_3 = x_1x_2x_3.$$ Then $x_1(x_2x_3-1) = 2x_2+3x_3$ and $x_1 = \dfrac{2x_2+3x_3}{x_2x_3-1}$ and I didn't see how to choose the $x_i$ here.

Answer

For $n=3$ you have $x_1 = \dfrac{2x_2+3x_3}{x_2x_3-1}$. You want to make the division come out even, and an easy way to do so is choose $x_2=2,x_3=1$ so the denominator is $1$. Now you can just compute that $x_1=7$. The same approach works for higher $n$. You have $x_1 = \dfrac{2x_2+3x_3+\ldots +nx_n}{x_2x_3\ldots x_n-1}$. Choose $x_2=2, x_i=1$ for $3 \le i \le n$ and let $x_1$ come out what it may, which happens to be $1+\frac 12n(n+1)$