# Number TheoryProving operations of congruence modulo m

#### crypt50

##### New member
If a, b and m > 0 are integers such that a % b (mod m), then a^n % b^n (mod m) for all positive integers n. I don't know how to go about it, any help would be greatly appreciated.

#### Ackbach

##### Indicium Physicus
Staff member
By '%', do you mean congruent? That's typically written
$$a \equiv b \;( \text{mod} \; m),\qquad \text{and} \qquad a^{n} \equiv b^{n} \;( \text{mod} \; m).$$
Use induction on $n$ to prove this. What will you need to show the inductive step?

#### Klaas van Aarsen

##### MHB Seeker
Staff member
Welcome to MHB, crypt50!

Assuming you meant what Ackbach suggested, here's an alternative way.

The expression $a \equiv b \pmod m$ means that there is a $k \in \mathbb Z$ such that $a=b+km$.
This implies that $a^n=(b+km)^n$.
Can you expand the right hand side with the binomial theorem?
If so, what can you conclude?