parmenides51 04.04.2021 20:14 Are there natural numbers $(m,n,k)$ that satisfy the equation $m^m+ n^n=k^k$ ?
Steve12345 04.04.2021 20:52 No. WLOG $k>m \ge n$. We have $(m+1)^{m+1} \le k^k \le 2m^m$ which can't be. Since $2 \ge \left( \frac{m+1}{m} \right)^m \cdot (m+1) > m+1 \ge 2$ .