Competition problem — Number theory

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Number theory Difficulty 5.9 AIME, harder Prove it

37. (ROM 3) Let $A_{1}, A_{2}, \ldots, A_{n+1}$ be positive integers such that $\left(A_{i}, A_{n+1}\right) =1$ for every $i=1,2, \ldots, n$ . Show that the equation
$x_{1}^{A_{1}}+x_{2}^{A_{2}}+\cdots+x_{n}^{A_{n}}=x_{n+1}^{A_{n+1}}$
has an infinite set of solutions $\left(x_{1}, x_{2}, \ldots, x_{n+1}\right)$ in positive integers.

Solution

37. We look for a solution with $x_{1}^{A_{1}}=\cdots=x_{n}^{A_{n}}=n^{A_{1} A_{2} \cdots A_{n} x}$ and $x_{n+1}=$ $n^{y}$ . In order for this to be a solution we must have $A_{1} A_{2} \cdots A_{n} x+1=$ $A_{n+1} y$ . This equation has infinitely many solutions $(x, y)$ in $\mathbb{N}$ , since $A_{1} A_{2} \cdots A_{n}$ and $A_{n+1}$ are coprime.

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Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic and difficulty added by this site.