Prove $2^k+1$ divisible by 3 for odd K

Question

Prove $2^k+1$ divisible by 3 iff $k$ is odd number.

Since I need to prove both direction looks like if I need to prove it's divisible by 3 it's by induction and the other side by congruence..am I right? is there a better wat than induction? and we I still can't prove both direction... any hints?

Try phrasing the question in terms of modular arithmetic rather than divisibility. — , Apr 03 '13 at 21:59

score 1 · Answer 1 · answered Apr 03 '13 at 21:59

1

HINT: Prove by induction that $2^k\equiv(-1)^k\pmod 3$.

answered Apr 03 '13 at 21:59

Brian M. Scott

616,228

Actually, induction is unnecessary. It suffices to consider that $2 \equiv -1$. – A.S Apr 04 '13 at 18:01
@Andrew: Depends on what the OP already knows about modular arithmetic. – Brian M. Scott Apr 04 '13 at 18:25
@A.S Actuallu induction is used in such modular proofs, but it is simply hidden in some lemma (here in the proof of the Congrence Power Rule $, a\equiv b,\Rightarrow, a^n\equiv b^n,,$ for $,a,b = 2,, -1).\ \ $ – Bill Dubuque May 03 '19 at 15:13

score 1 · Answer 2 · answered Apr 03 '13 at 21:59

1

Hint: $4 \equiv 1 \mod 3$, so $2^{k+2} \equiv \ldots $

answered Apr 03 '13 at 21:59

Robert Israel

448,999

score 1 · Accepted Answer · answered Apr 03 '13 at 22:00

1

If $k=2m$, then $2^k=4^m\equiv 1^m\pmod 3$ and if $k=2m+1$ then $2^k=2\cdot 4^m\equiv 2\cdot 1^m\pmod 3$.

answered Apr 03 '13 at 22:00

Hagen von Eitzen

374,180

so the other way is in induction as I said? – Mary Apr 04 '13 at 00:15
@Hagen von Eitzen can I just ask for some clarification, as I am learning this subject: I read the last part as $ 2^k \equiv 2 (mod 3) $ which by definition means $ 3|(2^k - 2)$ - so how is this the same as $ 3|(2^k + 1)$? Many Thanks. – JackReacher Aug 13 '13 at 02:18

score 0 · Answer 4 · answered Apr 03 '13 at 22:15

0

Hint $\rm\ \ 1\!-\!(-2)\mid 1\!-\!(-2)^{2n+1}$

answered Apr 03 '13 at 22:15

Math Gems

19,574

Prove $2^k+1$ divisible by 3 for odd K

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