# Thread: Proof of a congruence

1. ## Proof of a congruence

Prove (w/o induction) that B^n and 0 are congruent mod B.
Hint from professor: let n = m + 1

2. Originally Posted by jzellt
Prove (w/o induction) that B^n and 0 are congruent mod B.
Hint from professor: let n = m + 1
Hint isn't really necessary as long as $n \ge 1$, which i think you are assuming

note that $B^n \equiv 0 ~(\bmod ~B) \Longleftrightarrow B \mid (B^n - 0)$

this is obviously the case.

So begin by saying $B^n = B^{m + 1} = B \cdot B^m$

where can we go with this?

3. I guess I should've post this in my post...but I actually got that far.

Since B^n = B^(m+1) then B|(BB^m) - 0 or B|BB^m.

But, Im not sure what to do from here.

4. Originally Posted by jzellt
I guess I should've post this in my post...but I actually got that far.

Since B^n = B^(m+1) then B|(BB^m) - 0 or B|BB^m.

But, Im not sure what to do from here.
since $B^n = B \cdot B^m$ we have that $B | B^n \Longleftrightarrow B|(B^n - 0)$ which means?

5. You did say without induction this time, didn't you?

B is congruent to 0, mod n, if and only B is a multiple of n: say B= na where a is some integer.

If B is congrunt to 0 mod n, then B= na. $B^m= (na)^B= n^ma^m= n(n^{m-1}a^n)$ is sufficient.

And, of course, 0= 0n.