a problem in elementary number theory :$n=a+b$ how to prove if $n\in \Bbb N $and $n\gt 6$ then there exists $a,b \in \Bbb N $ such that $ a,b\ge 2$ and $\gcd(a,b)=1 $ and $n=a+b$.
Thanks in advance. 
 A: Take one of $a,\ b$ to be the smallest prime $p$ with $p\nmid n$ (then why $a,b\neq1?$). 
A: If $\rm\:n = a\!+\!b\:$ then $\rm\: 1=(a,b)=(a,n\!-\!a) = (a,n).\,$ Note $\rm\,a,b\ge 2$ $\iff$ $\rm a\ne 1,\ a\ne n\!-\!1.\,$ Thus we need to prove that there is some $\rm\ a\in[2,\,n\!-\!2]\ $ such that $\rm\ a\ $ is coprime to $\rm\,n,\ $ i.e. $ $  that $\rm\,phi(n) > 2.\ $ If you know Euler's phi formula then this is simple.
Otherwise, if $\rm\,n>3\,$ is odd, choose $\rm\,a=2;\,$ else $\rm\,n\,$ is even, so $\rm\,n = 2^k m,\,$ for odd $\rm\,m.\:$ Show that $\rm\ a = 2\!+\!m\ $ works for $\rm\,n>6.$
Remark $\ $ Above we constructed a smaller natural $\rm\,2\!+\!m\,$ coprime to $\rm\:2^k m\:$ by splitting it into coprime factors, then adding them (or their radicals). This is the same construction used in variant of Euclid's proof that there are infinitely many primes. This uses $\rm\:(n,m)=1\:\Rightarrow\:(nm,\,n\!+\!m) = 1\:$. Thus any factorization $\rm\:k = nm\:$ into coprime factors yields a number $\rm\,n\!+\!m\,$ coprime to $\rm\,k,\,$ and which, generally, is smaller than $\rm\,k\,$ (as we need above). Being coprime to $\rm\,k\,$ and $> 1$ it has a prime factor not in $\rm\,k,\,$ hence it yields a "new" prime.
Ribenboim attributes this form of Euclid's proof to Stieltjes (1890) (but I would not be surprised if it is much older). Note that Euclid's proof is a special case, using a trivial factorization, i.e. $\rm\,m = 1.$
A proof that $\rm\,(n,m)=1\,\Rightarrow\,(nm,n\!+\!m)=1\,$ is easy. First $\rm\,1 = (n,m) = (n,n\!+\!m).\, $  Similarly $\rm\,(m,n\!+\!m)=1,\:$ thus $\rm\,(nm,n+m)=1\:$ follows by a form of Euclid's Lemma: if both $\rm\,n,m\,$ are coprime to an integer, then so too is their product. This coprimality often proves handy.
