## Congruence of all integers n, 4^n and 1 +3n mod(9)?

I just took a number theory midterm, the professor had a question the that said
"Show by induction that for all integers n, 4$$^{n}$$ is congruent to 1 +3n mod(9).

Now am I crazy or did the professor probably mean to say integers greater or equal to 0, or for any natural number n, ...

couldn't you show a counter example for instance n = -2, such that the congruence is false?
 PhysOrg.com science news on PhysOrg.com >> City-life changes blackbird personalities, study shows>> Origins of 'The Hoff' crab revealed (w/ Video)>> Older males make better fathers: Mature male beetles work harder, care less about female infidelity

Recognitions:
Homework Help
 Quote by mgiddy911 I just took a number theory midterm, the professor had a question the that said "Show by induction that for all integers n, 4$$^{n}$$ is congruent to 1 +3n mod(9). Now am I crazy or did the professor probably mean to say integers greater or equal to 0, or for any natural number n, ... couldn't you show a counter example for instance n = -2, such that the congruence is false?
In number theory, it's not unusual to assume that variables are natural numbers unless otherwise specified. I don't really want to think about negative powers here, since:
* There is no multiplicative inverse over natural numbers or integers
* There is a multiplicative inverse over real numbers (x not 0)
* There is sometimes a multiplicative inverse over Zp

and there could be reason to work over any of these.
 Recognitions: Gold Member The only residues are 1,4,7, so even if we use the inverses, it doesn't matter since $$\frac{1}{4}\equiv 7 mod 9$$

Recognitions:
Gold Member

## Congruence of all integers n, 4^n and 1 +3n mod(9)?

When you think about it, as a general rule, if a and Mod b are such that, a, b positive integers, and (a,b) =1, (they are relatively prime ) Then the powers of a Mod b form a cyclic group, i.e., every power of a has an inverse in the group.