3^k not congruent to -1 mod 2^e, e > 2.

## Primary tabs

# 3^k not congruent to -1 mod 2^e, e > 2.

Submitted by estefan34 on Thu, 04/12/2012 - 07:54

Forums:

$3^k \not\equiv -1 $ mod $ 2^e $ for e > 2, k > 0. Is this true? I have tried to prove it by expanding (1 + 2)^k. [Notation: (n; m) := n! / (m! (n - m)!)] E.g., for e = 3 I get: (1+2)^k + 1 = 2 + (k; 1) 2 + (k; 2) 2^2 + (k; e) 2^e + ... So, here it's enough to prove that 2^3 does not divide 2 + (k; 1) 2 + (k; 2) 2^2. The validity for general e seems very hard to prove.

- Forums
- Planetary Bugs
- HS/Secondary
- University/Tertiary
- Graduate/Advanced
- Industry/Practice
- Research Topics
- LaTeX help
- Math Comptetitions
- Math History
- Math Humor
- PlanetMath Comments
- PlanetMath System Updates and News
- PlanetMath help
- PlanetMath.ORG
- Strategic Communications Development
- The Math Pub
- Testing messages (ignore)

- Other useful stuff
- Corrections

## Re: 3^k not congruent to -1 mod 2^e, e > 2.

It was not so hard. Having proved that the proposition is true for e =3, I have that 2^3 does not divide 3^k + 1 for k any natural number. But then, all the more so 2^e does not divide 3^k + 1 for e > 3. I've made a fool of myself!