# Proof about sum of integers

1. Jun 20, 2013

### cragar

1. The problem statement, all variables and given/known data
Prove that every integer bigger than 6 can be written as a sum of 2 integers
bigger than 1 which are relatively prime.
3. 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 Lets 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. Lets assume for contradiction that n and$n+2^x$ 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 lets 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.

2. Jun 20, 2013

### lurflurf

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