 That is, we may "divide by "b"" when working modulo "a".Furthermore, if "b"1 and "b"2 are both coprime with "a", then so is their product "b"1"b"2 (i.e., modulo "a" it is a product of invertible elements, and therefore invertible); this also follows from the first point by Euclid's lemma, which states that if a prime number "p" divides a product "bc", then "p" divides at least one of the factors "b", "c".As a consequence of the first point, if "a" and "b" are coprime, then so are any powers "a""k" and "b""m".