Proving divisibility with consecutive integers

Consecutive integers cycle through all remainders for the length of their block. In any three consecutive integers, one is divisible by three; in any two consecutive integers, one is even. Those facts can prove divisibility without expanding a product. State that the variable is an integer, because the conclusion fails for arbitrary real inputs.

Try small integers to look for a pattern, then prove it using a general integer. When combining divisibility statements, consider the factors involved. Being divisible by both two and three implies divisibility by six because those factors are coprime. Being divisible by both two and four does not by itself imply divisibility by eight. A proof must explain the specific factors needed.

Worked example

Prove that n(n + 1)(n + 2) is divisible by 6 for every integer n.

1. Among n and n + 1, at least one is even, so the product of all three integers is divisible by 2.
2. The three consecutive integers represent all three remainders modulo 3, so one is divisible by 3.
3. Since 2 and 3 have no common factor greater than 1, a product divisible by both is divisible by 6.

Answer: n(n + 1)(n + 2) = 6k for some integer k, for every integer n.

Revise first: Algebraic proof and counterexamples.