$2^{2n+1} +1$ divisible by $3$

Question

$2^{2n+1} +1$ divisible by $3$

Inducción over n

Let $n=1$, then: $$2^{2(1)+1}+1=2^{3}+1=9$$ It works!

Hip. $$2^{2n+1} +1=3k$$

Then we have to show: $$2^{2n+3} +1=3k$$

Any guess?

Answer 1

Another idea:

$$2^{2n+3}+1=4\cdot2^{2n+1}+1=\overbrace{2^{2n+1}+1}^{=3k,\,\text{by Ind Hyp.}}+\color{red}3\cdot2^{2n+1}$$

Answer 2

HINT

$$2^{2n+3} +1=2^2 \cdot 2^{2n+1} +1\stackrel{Hyp.}=2^2(3k-1)+1$$

Answer 3

Hint:

$$2^{2n+3}+1=(2^{2n+3}-2^{2n+1})+(2^{2n+1}+1).$$

Answer 4

Conceptually the induction follows very simply by multiplying the congruences below using CPR = Congruence Product Rule $ $ as follows $$\begin{align}\bmod 3\!:\qquad \color{#c00}{2^{\large 2}}\ &\equiv\ \ \ \color{#c00}{1}\\ 2^{\large 2n+1}&\equiv -1\qquad\ \ P(n)\\ \Rightarrow\ \ \color{#c00}{2^{\large 2}}\,2^{\large 2n+1}&\equiv -1\cdot \color{#c00}{1}\quad\ P(n\!+\!1)\end{align}\quad\ \ \ $$

See this answer for further discussion, including how to do the above without congruences.