3-ча 1. Solve the equation \(x^{y}=y^{x}(x \neq y)\) in natural numbers. Find all solutions.
Solve 1. Answer: \(x=2, y=4\) or \(x=4, y=2\). From the equality \(x^{y}=y^{x}\) it follows that the prime divisors of the numbers \(x\) and \(y\) are the same, i.e., \(x=p_{1}^{a_{1}} p_{2}^{a_{2}} \ldots p_{n}^{a_{n}}\) and \(y=p_{1}^{b_{1}} p_{2}^{b_{2}} \ldots p_{n}^{n_{n}}\). The same equality shows that \(a_{1} y=b_{1} x, \ldots, a_{n} y=b_{n} x\). Let for definiteness \(x<x\), therefore \(k>1\), and thus \(x>1\). It is clear that \(2^{2-1}=2\). It is also easy to verify that if \(x>2\) or \(k>2\), then \(x^{k-1}>k\).