Originally Posted by

**Soroban** Hello, yuud!

Here is a sneaky method . . . if you know modulo arithmetic.

$\displaystyle 3^{2n+1} + 2^{n+2} \;\;=\;\left(3^3\right)^\frac{2n+1}{3}} + \left(2^3\right)^{\frac{n+2}{3}} $

. . . . . . . . . . .$\displaystyle =\;(27)^{\frac{2n+1}{3}} + (8)^{\frac{n+2}{2}} $

. . . . . . . . . . .$\displaystyle \equiv\; (\text{-}1)^{\frac{2n+1}{3}} + (1)^{\frac{n+2}{2}}\text{ (mod 7)}$

. . . . . . . . . . .$\displaystyle \equiv\;\left(\sqrt[3]{\text{-}1}\right)^{2n+1} + \left(\sqrt[3]{1}\right)^{n+2}\text{ (mod 7)}$

. . . . . . . . . . .$\displaystyle \equiv\; (\text{-}1)^{2n+1} + (1)^{n+2}\text{ (mod 7)} $

. . . . . . . . . . .$\displaystyle \equiv\; -1 + 1\text{ (mod 7)}$

$\displaystyle 3^{2n+1} + 2^{2n+1} \;\equiv\;0 \text{ (mod 7)}$

It is divisible by 7 for *all* positive intergers $\displaystyle n$.