Show that a positive integer $n \in N$ can be written as a sum of positive coprime integers with $gcd(a,b)=1 $

Question

My idea was to show this via 3 cases. In case one n is even n=2k, k is odd In case two n is even with n=2k, k is odd In case three n is odd so n=2k+1 Then I have to show that for $n<7$ not every integer has a written form of $n=a+b; gcd(a,b)=1.$ I want to show that in each case, the $gcd(a,b)=1$. Is this the right approach ?

Answer 1

If one of the numbers is allowed to be $1$, then $a=n-1$ and $b=1$ is always a solution. If not, $n$ must be at least $7$.

If $n \geq 7$ is odd, then $n=2k+1 = k+(k+1)$ for some integer $k \geq 3$. Consecutive integers are always relatively prime, so $a=k$ and $b=k+1$ would work.

If $n$ is even, you have to try something else. You could write $n$ as the sum of the two integers which differ by $2$, but they may not be relatively prime. For instance, $10=4+6$. But $12=5+7$. The difference is the congruence modulo $4$.

If $n\geq 7$ is a multiple of $4$, then $n=4k = (2k-1) + (2k+1)$ for some $k\geq 2$. That is, $n$ is the some of consecutive odd integers. By Euclid's algorithm, $$ \gcd(2k+1,2k-1) = \gcd(2k-1,2) = 1 $$ So $a=2k-1$ and $b=2k+1$ will work.

If $n\geq7$ is even, but not a multiple of $4$, then $n= 4k + 2 = (2k-1) + (2k+3)$ for some $k\geq2$. As above, $$ \gcd(2k+3,2k-1) = \gcd(2k-1,4) = 1 $$ So $a=2k-1$ and $b=2k+3$ will work. In the case of $n=22$, for instance, you get $a=9$ and $b=13$.

Here is a less case-based and less constructive solution: Let $p$ be a prime less than $n$ which does not divide $n$. Let $a=b$ and $b=n-p$. Then $$ \gcd(n-p,p) = \gcd(n,p) = 1 $$ Since $p \nmid n$. In the case of $n=22$, you get $3+19$, $5+17$, $7+15$, and $13+9$.