Prove that every natural number $n >6$ can be written as a sum $a+b$ where $a, b∈\mathbb N\setminus\{1\}$ and$ \gcd(a, b) = 1$

Question

Is this the correct approach to this problem: I tried induction. This is what the professor said: Hint: Treat the cases $n$ even and odd separately.

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Inductive hypothesis:

$n=a+b$ for $n>6$

Base case

Let $n = 7$. Then $3+4=7$ where $3,4$ have a gccd of $1$.

Induction Step

Induction Step: Prove: $n+1=a+b$

Case 1: Let $a$ be $a + 1$. This is possible because $a$ is in the natural numbers. So $a+b+1=n+1$. By the inductive hypothesis $n=a+b$ so we can substitute it. Therefore $n+1=n+1$.

Case 2: Let $b$ be $b+1$. This is symmetrical to case one.

Answer 1

Actually induction is not needed. Assuming $n>6$ we have that $\varphi(n)$ is even and $\geq 4$.
Once we pick some $a\in[1,n-1]$ which is coprime with $n$ we also have that $b=n-a$ is coprime with $n$, and $a,b$ cannot have common divisors except $1$, since a non-trivial common divisor of $a,b$ would be a non-trivial common divisor of $a,n$.

In particular, it is enough to pick $a$ as the smallest prime $\nmid n$ and $b$ as $n-a$.