Consecutive odd natural numbers - one is composite. Prove

[nelson98]

Homework Statement

Every triple of consecutive odd natural numbers, with the first being at least 5, contains at least on composite.

Homework Equations

N/A

The Attempt at a Solution

I know from number theory that of every set of consecutive odd integers, one of them is divisible by three, thereby making it a composite number. I just don't know how to prove it. Can anybody help?

 

[HallsofIvy]

A triple of consectutive odd integers can be written as 2n+1, 2n+ 3, and 2n+ 5.

Suppose 2n+ 1 is NOT a multiple of 3. Then it has a remainder of either 1 or 2 when divided by 3

a) Suppose 2n+ 1 has remainder 1 when divided by 3: 2n+ 1= 3k+ 1 so that 2n= 3k. Look at both 2n+3 and 2n+ 5 in this case.

b) Suppose 2n+ 1 has remainder 2 when divided by 3: 2n+ 1= 3k+ 2 so that 2n= 3k+ 1. Look at both 2n+3 and 2n+ 5 in this case.