# Congruent Integers

1. Feb 18, 2010

### scottstapp

1. The problem statement, all variables and given/known data
I need to prove the following but have no idea how to do so.

Let a,b, k be integers with k positive. If a is congruent to b(mod n), then ak is congruent to bk (mod n).

2. Relevant equations
The hint given is that I can assume the following proposition is true and that I am supposed to use it to show the statement holds for k=2,3...

Proposition:
If a is congruent to b(mod n) and c is congruent to d(mod n) then a+c is congruent to b+d(mod n)

Thanks for your help, I am pretty lost on this so anything helps.

2. Feb 19, 2010

### Staff: Mentor

The directions of this problem seem to be pointing you to doing it by induction on k. For k = 1, the proposition is obviously true.

Assume the proposition is true for k = m. IOW, assume that
am $\equiv$ bm mod n.

Now use this assumption to show that the proposition is true for k = m + 1. I.e., that
am + 1 $\equiv$ bm + 1 mod n.

Something that might be helpful is that if p $\equiv$ q mod n, then p - q $\equiv$ 0 mod n.

3. Feb 19, 2010

### HallsofIvy

Staff Emeritus
Simpler, I think. $a^k- b^k= (a- b)(a^{k-1}+ a^{k-2}b+ \cdot\cdot\cdot ab^{k-2}+ b^{k-1})$. Now use the fact that, since a= b (mod n), a- b is a multiple of n.

4. Feb 19, 2010

### Staff: Mentor

That's similar to the direction I was sending the OP, but didn't want to get dinged again for giving too much help...