1. ## prove ....

prove that if n>6 then $\displaystyle \exists a,b \in N \ a,b >1 , \ gcd (a,b)=1$ such that n= a+b

2. Originally Posted by flower3
prove that if n>6 then $\displaystyle \exists a,b \in N \ a,b >1 , \ gcd (a,b)=1$ such that n= a+b

If n is odd you can take $\displaystyle a=n-2\,,\,\,b=2$ . I'll let you the case for even n. Can you say now why they required $\displaystyle n>6$ ?

Tonio

3. Hello,
$\displaystyle n > 6$. Consider two cases :

$\displaystyle \Rightarrow n$ is odd. Let $\displaystyle a = \frac{n - 1}{2}$ and $\displaystyle b = \frac{n + 1}{2}$. $\displaystyle a + b = \frac{2n}{2} = n$, and $\displaystyle b - a = \frac{n + 1 - n + 1}{2} = 1$. Since $\displaystyle b = a + 1$, $\displaystyle gcd(a, b) = 1$.

$\displaystyle \Rightarrow n$ is even. Let $\displaystyle a = \frac{n - \beta}{2}$ and $\displaystyle b = \frac{n + \beta}{2}$, for some $\displaystyle \beta$ odd prime. $\displaystyle a + b = \frac{2n}{2} = n$, and $\displaystyle b - a = \frac{n + \beta - n + \beta}{2} = \beta$. Since $\displaystyle b = a + \beta$, $\displaystyle gcd(a, b) = 1$ if and only if $\displaystyle gcd(a, \beta) = 1$. Since a number cannot potentially divide every prime less than itself (otherwise it would at least be the product of all these primes and would be much greater, leaving more primes in between), there must exist some $\displaystyle \beta$ for which $\displaystyle gcd(a, \beta) = 1$, and $\displaystyle gcd(a, b) = 1$ follows.

As you can see, the case for $\displaystyle n$ even is much harder than the case for $\displaystyle n$ odd. Can you see how I used divisibility theorems and properties of prime numbers to solve this problem ? If you have any questions/concerns, feel free to ask.