Question 870785
If p | a, we are done.


Otherwise, p and a are relatively prime, so by Bezout's identity, there exist integers m and n such that pm + an = 1. Multiplying both sides by b, pmb + abn = b. Since pmb and abn are divisible by p (p | ab), then the sum pmb + abn is divisible by p, so b is divisible by p.