prove that $2^{n}+1$ is divisible by $n=3^k$ for $k≥1$ [closed]

prove that :

$$2^n+1$$ is divisible by all number from : $$n=3^k$$

for $$k≥1$$

I find this problems in book and I need ideas to approach it

Problems : closed as off-topic by Morgan Rodgers, Dietrich Burde, Nosrati, Leucippus, GAVDMay 15 at 4:12

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• Do you mean to say: "if $n=3^k$, then $2^n+1$ is divisible by $n$"? – kccu May 14 at 17:54
• Yes sir , actually I'm not good in English – Kînan Jœd May 14 at 18:05
• Not everyone here is a sir. – kccu May 14 at 20:06

Approach by induction. We first check that it is true for $$k=1$$. Indeed, $$2^{(3^1)}+1=9$$ is divisible by $$3^1=3$$

Rewording the claim, $$3^k \mid 2^{(3^k)} + 1$$. Reworded again, there exists some $$a$$ such that $$2^{(3^k)}+1 = 3^k\cdot a$$

Suppose that the claim is true for some $$k\geq 1$$. We try to show that it is also true for $$k+1$$.

$$2^{(3^{k+1})}+1 = 2^{3(3^k)}+1 = (2^{(3^k)})^3+1 = (3^k\cdot a - 1)^3+1$$

$$=(3^k)^3\cdot a^3-3\cdot (3^k)^2\cdot a^2+3\cdot 3^k\cdot a - 1 + 1$$

Now, from here, it should be clear that after cancelling the $$-1$$ and the $$+1$$ that each term is divisible by $$3\cdot 3^k$$ and the claim is proven.

• Thank you very much – Kînan Jœd May 14 at 18:06