Moonmathpi496 wrote: Find the smallest natural number $ k$ for which there exists natural numbers $ m$ and $ n$ such that $ 1324 + 279m + 5^n$ is $ k$-th power of some natural number. Trivially $ k>1$, for $ k=2$ taking mod $ 9$ $ 1324\equiv 1(9)$, $ 279 m \equiv 0 (9)$ and $ 5^n \equiv 5, 7, 8, 4, 2, 1 (9)$ for $ n=1,2,3,4,5,6$ The quadratic residues mod $ 9$ are: $ 0,1,4,7$ therefore $ n=3+6k=3(1+2k)$ taking mod $ 31$ $ 1324\equiv 22(31)$, $ 279 m \equiv 0 (31)$ and $ 5^n \equiv 5,25,1 (31)$ for $ n=1,2,3$ but $ 23$ nor $ 27$ are quadratic residues mod $ 31$ therefore $ n=2+3j$ contradiction For $ \boxed{k=3}$ $ 12^3=1324+279\cdot 1+5^3$