Odd power of 2 plus even power of 2, is divisible by 6

Question

Prove $2^{2m+1}+2^{2n}$ is divisible by 6 for all integers $n\geq1$ and $m\geq0$. Equivalently one can prove $2+4^k$ is divisible by 3.

Answer 1

$$2^{2m+1}+2^{2n}=2\cdot4^m+4^n\equiv 2\cdot1^m+1^n\equiv2+1\equiv 0\pmod3$$

And the number is obviously even, so it is divisible by 6.

Answer 2

Hint

$$2+4^k=3+ \left( 4^k-1 \right)=3+ (4-1) \left(4^{k-1}+...+4+1 \right)$$

Answer 3

We have $2\equiv -1 \bmod 3$ which trivially makes the sum $\equiv 0 \bmod 3$ (odd power plus even power).

Answer 4

Quite obvious with congruences modulo $6$:

It is easy to check that $\enspace2^n\equiv\begin{cases}2&\text{if $n$ is odd,}\\ 4&\text{if $n$ is even.} \end{cases}$

Thus the sum of an even and an odd power of $2$ is congruent to $4+2\equiv 0\mod 6$.

Still more obvious for the equivalent assertion: $4\equiv 1\mod 3$, so $$2+4^k\equiv 2+1^k=2+1\equiv 0\mod 3.$$