Prove that $2^n +1$ is divisible by $3$ for all positive integers $n$.

Question

I just want to know if I went on the right direction. With induction

Let $n=1$, then $2^1+1= 3$, which is divisible by $3$. Then show proof for $n+1.$

$2^n+1=3k$

So we get $2^{n+1}+1, \rightarrow 2^n+2+1, \rightarrow 3k+3= 3(k+1)$. Thus $2^n+1$ is divisible by $3$.

Now if I wanted to show that $2^n+1$ is divisble by $3$, $\forall$ odd integers $n$. Would it be with induction:

$n=1$, then $2+1=3$, and $3|3$. Let $n=2k+1$, since n is odd, then we get $2^{2k+1}+1=3m$. Now we need to show for $k+1$.

We get: $2^{2k+2+1}+1=3m \rightarrow 2^{2k+1}*2^2+4-3 \rightarrow 4(2^{2k+1}+1)-3$

$\rightarrow 4(3m)-3 \rightarrow 3(4m-1)$, thus $2^n+1$ is divisible by $3$.

$2^2+1=5$ is not divisible by $3$. You mean odd integers $n$. — J.R., Oct 16 '17 at 15:27
$4$th line is wrong. $2^{n+1}=2^n\cdot 2$ and not $2^n +2$ as you wrote — Raffaele, Oct 16 '17 at 15:28
By FT= Factor Theorem $,3=2-(-1)\mid 2^n-(-1)^n, (= 2^n+1,$ by $n$ odd). See the dupe for many ways to prove FT (which can be specialized to direct proofs here if so desired, including all proofs below) $\ \ $ — Bill Dubuque, Mar 19 '24 at 19:11

score 5 · Answer 1 · answered Oct 16 '17 at 15:27

5

But it is not true, say $n=2$. :(

If $n$ is odd then $2^n+1 = (2+1)(2^{n-1}-...+1) =3k$, then is true.

answered Oct 16 '17 at 15:27

nonuser

Try $n=-1$. I think $-1$ is an odd integer number. – Michael Rozenberg Oct 16 '17 at 15:53
In that case $2^n+1$ it is not even an integer – nonuser Oct 16 '17 at 15:54
If so by your way for $n=2$ we get: $2^2+1=5=3k$, where $k=\frac{5}{3}$. – Michael Rozenberg Oct 16 '17 at 16:00
Which way...? What are you trying to prove Michael? – nonuser Oct 16 '17 at 16:02

score 4 · Accepted Answer · answered Oct 16 '17 at 15:29

4

answered Oct 16 '17 at 15:29

score 3 · Answer 3 · answered Oct 16 '17 at 15:29

3

This is only true for odd integers $n=2k+1$:

$$2^{2k+1}+1 = 2^{2k+1}-(-1)^{2k+1} = (2+1) (2^{2k}-2^{2k-1}\pm\cdots+1)$$

answered Oct 16 '17 at 15:29

J.R.

score 3 · Answer 4 · answered Oct 16 '17 at 15:41

If you need to prove it for odd $n$, then here is an elegant proof

$$2^n=(3-1)^n$$ $$=3k+(-1)^n$$ Since $n$ is odd , $(-1)^n=-1$

Also proving $2^n+1$ is divisible by $3$ is same as proving $2^n-2$ is divisible by $3$

Therefore $$ 2^n-2= (3-1)^n-2$$ $$=3k+(-1)^n-2$$ $$=3k-1-2$$ $$3k-3 $$ And this is divisible by $3$ for all $n\in \mathbb N(odd)$

And of course it is not true for every even $n$

score 2 · Answer 5 · answered Oct 16 '17 at 15:37

2

$2^n+1$ is divisible by $3$ only when $n = 2k+1, (k∈\mathbb{Z}^*)$.

Use modulus:

For every odd number $n$, we have $$2^n≡2 (\mod n)$$ and for every even number $n$, we have $$2^n≡1 (\mod n)$$

answered Oct 16 '17 at 15:37

iBug

score 0 · Answer 6 · answered Aug 16 '22 at 11:25

0

We know that for all n = odd positive integers, a+b|a^n+b^n Therefore, 3 i.e. 2+1|2^n+1^n for n is odd positive integer

answered Aug 16 '22 at 11:25

Yash Chauhan

6 Answers6