show that
37^n+2 + 16^n+1 + 23^n is divisible by 7 whenever n is natural
thank you
I assume you're having trouble with step 3.
Then note:
$\displaystyle 37^{k+3} + 16^{k+2} + 23^{k+1} = 37 \cdot 37^{k+2} + 16 \cdot 16^{k+1} + 23 \cdot 23^{k}$
$\displaystyle = 37 \left( 37^{k+2} + 16^{k+1} + 23^{k} \right) - 21 \cdot 16^{k+1} - 14 \cdot 23^k$
$\displaystyle = 37 \left( 37^{k+2} + 16^{k+1} + 23^{k} \right) - 7 \left(3 \cdot 16^{k+1} + 2 \cdot 23^k\right)$
which is divisible by 7 due to step 2 ........
Here is a non-induction proof.
Let $\displaystyle A_n = 37^{n+2}+16^{n+1}+23^n$.
Note $\displaystyle 37^{n+2}\equiv 2^{n+2} (\bmod 7)$, $\displaystyle 16^{n+1}\equiv 2^{n+1}(\bmod 7)$, $\displaystyle 23^n\equiv 2^n(\bmod 7)$.
Thus, $\displaystyle A_n \equiv 2^{n+2}+2^{n+1}+2^n = 2^n(1+2+2^2) \equiv 0 (\bmod 7)$.