Proof: If a ≡ b (mod n) then am ≡ bm (mod mn) and converse

Question

Prove the following, for positive integers m and n.

If $a \equiv b \pmod n$ then $am \equiv bm \pmod{mn}$ and conversely, If $am \equiv bm \pmod{mn}$ then $a \equiv b \pmod n$.

This makes sense to me logically. And I believe that this has something to do with the definition of multiples or divisors in modular arithmetic. Since $a - b$ is a multiple of $n$. Something along the lines of then also $am - bm $would be a multiple of $mn$. I'm not sure if the steps would be reversible so that I could also prove the converse.

However, I do not not know how to formalize this proof, or starting steps. Thank you for any help and support.

Answer 1

$a ≡ b\mod n$ ⇒ $a=k.n +b$ ⇒ $ am=k(nm)+mb$ ⇒ $am≡mb \mod nm$

Do reverse operation for reverse:

$am≡mb \mod nm$ ⇒ $ am=k(nm)+mb$ ⇒ $a=k.n +b$ ⇒$a ≡ b\mod n$

Answer 2

if $a-b$ is a multiple of $n$, it means there exists $k \in \mathbb{Z}$ such that $a-b = kn$, you just have to multiply by $m$ to get the result.

Hint:

To solve the converse. think of what is the reverse operation of multiplication. Note that $m$ is positive.

Answer 3

Part 1, If a ≡ b (mod n) then we can say that n | a - b. So a - b = kn (k ∈ Z) Multiplying m in both sides we get, am - bm = mnk So, mn | am - bm. This implies, a ≡ b (mod mn) Part 2 if am ≡ bm (mod mn), (am - bm) = mnx (x ∈ Z) or, m(a - b) = mnx or, (a - b) = nx This implies, n | (a - b) or a ≡ b (mod n) Proved!