3-ча 1. Find all positive rational solutions of the equation \(x^{y}=y^{x}(x \neq y)\).
Solution 1. Let \(y=k x\). Then \(x^{k x}=(k x)^{x}\). Extracting the root of degree \(x\), and then dividing by \(x\), we get \(x^{k-1}=k\). Thus, \(x=k^{\frac{1}{k-1}}\) and \(y=k k^{\frac{1}{k-1}}=k^{\frac{k}{k-1}}\). The number \(k\) is rational; let \(\frac{1}{k-1}=\frac{p}{q}\) be an irreducible fraction. Substituting this expression into the formulas for \(x\) and \(y\), we get \(x=\left(\frac{p+q}{p}\right)^{\frac{p}{q}}\) and \(y=\) \(\left(\frac{p+q}{p}\right)^{\frac{p+q}{q}}\). The numbers \(p\) and \(q\) are coprime, so the numbers \(x\) and \(y\) can only be rational if the integers \(p\) and \(p+q\) can have a root of degree \(q\) extracted from them. But if \(q \geqslant 2\) and \(p=n^{q}\), then

\[
n^{q}<p+q<(n+1)^{q}=n^{q}+q n^{q-1}+\frac{q(q-1)}{2} n^{q-2}+\ldots
\]

Therefore, \(q=1\), and thus \(x=\left(\frac{p+1}{p}\right)^{p}\) and \(y=\left(\frac{p+1}{p}\right)^{p+1}\); here \(p\) is any integer, except 0 and -1.