How can integers be written as sums of relatively prime integers?

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SUMMARY

Every integer greater than 6 can be expressed as a sum of two relatively prime integers greater than 1. The discussion outlines methods for both odd and even integers, demonstrating that for an odd integer x, the sum (x-2) + 2 is valid, while for an even integer 2y, the sum (y+1) + (y-1) holds true. Additionally, the lemma confirms that odd integers maintain relative primality with their adjacent even counterparts. The analysis includes specific cases for integers in the forms of 2n+1, 2n, and 4n.

PREREQUISITES
  • Understanding of basic number theory concepts, particularly relative primality.
  • Familiarity with integer properties, including odd and even classifications.
  • Knowledge of mathematical proofs and contradiction techniques.
  • Ability to manipulate algebraic expressions involving integers.
NEXT STEPS
  • Explore advanced number theory concepts, such as the Euclidean algorithm for finding greatest common divisors.
  • Study the properties of prime numbers and their role in relative primality.
  • Investigate other mathematical proofs involving sums of integers, such as Goldbach's conjecture.
  • Learn about combinatorial number theory and its applications in partitioning integers.
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Mathematicians, students studying number theory, educators teaching integer properties, and anyone interested in mathematical proofs related to relative primality.

cragar
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Homework Statement


Prove that every integer bigger than 6 can be written as a sum of 2 integers
bigger than 1 which are relatively prime.

The Attempt at a Solution


Ill first look at the case where our number is odd.
Let x be an odd integer. I will just add (x-2)+2=x since x is odd so is x-2 and 2 is even
so x-2 and 2 are relatively prime.

Now Let's look at the case where our number 2y is even.
and y is even. 2y=y+y=(y+1)+(y-1) now since y is even y+1 and y-1 are odd. and y-1 and y+1 are odd numbers separated by a factor of 2.
Lemma 1: Let n be an odd number. Let's assume for contradiction that n andn+2^x have a common factor so it should divide their difference but n+2^x-n=2^x but n and n+2^x do not have a factor of 2 because they are odd.
so y+1 and y-1 are relatively prime by lemma 1.Now let's look at the case where 2z=z+z where z is odd.
we will just look at 2z=z+z=(z+2)+(z-2) since z is odd z-2 and z+2 are odd and they are odd numbers separated by a power of 2 so they are relatively prime.
 
Physics news on Phys.org
Consider 3 cases
1)2n+1=(n)+(n+1)
any n
2)2n=(n+1)+(n-1)
n odd
3)4n=(2k+1)+(2k-1)
any n
 

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