Consecutive odd natural numbers - one is composite. Prove

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SUMMARY

The discussion centers on proving that every set of three consecutive odd natural numbers, starting from 5, contains at least one composite number. The participants highlight that among any three consecutive odd integers expressed as 2n+1, 2n+3, and 2n+5, at least one must be divisible by 3, thus ensuring its compositeness. The proof strategy involves analyzing the remainders when these numbers are divided by 3, confirming that one of them will always yield a composite result.

PREREQUISITES
  • Understanding of odd natural numbers and their properties
  • Basic knowledge of number theory concepts
  • Familiarity with divisibility rules, particularly for the number 3
  • Ability to manipulate algebraic expressions involving integers
NEXT STEPS
  • Study the properties of odd and even integers in number theory
  • Learn about divisibility tests and their applications
  • Explore proofs involving modular arithmetic
  • Investigate the characteristics of composite and prime numbers
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Mathematics students, educators, and anyone interested in number theory, particularly those focusing on properties of integers and proofs involving divisibility.

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Homework Statement



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

Homework Equations



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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?
 
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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.
 

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