Induction proof: $2^n + 3^n ≡ 5^n (mod 6)$

Once you check for $n=1$, you want to show: for $k\geq 1$, $$ 2^k + 3^k ≡ 5^k\pmod{6}\quad\text{implies}\quad 2^{k+1} + 3^{k+1} ≡ 5^{k+1}\pmod{6}. $$ So start with $2^k + 3^k ≡ 5^k\pmod{6}$. This implies we can write $5^k=2^k+3^k-6m $ for some integer $m$. Proceeding from here, we have \begin{align*} 5^{k+1}=5(2^k+3^k-6m)&=2^{k+1}+3\cdot2^k+3^{k+1}+2\cdot 3^k-30m\\ &=2^{k+1}+3^{k+1}+6(2^{k-1}+3^{k-1}-5m) \end{align*} which indeed implies $2^{k+1} + 3^{k+1} ≡ 5^{k+1}\pmod{6}$.