Integers as Sum of Three Pairwise Coprime Integers

Theorem
Let $n$ be an integer greater than $17$.

Then $n$ is the sum of $3$ integers greater than $1$ which are pairwise coprime.

Case $1$: $n$ is even
There is some integer $k > 2$ such that one of the following holds:


 * $n = 6 k = 2 + 3 + \paren {6 k - 5}$


 * $n = 6 k + 2 = 4 + 3 + \paren {6 k - 7}$


 * $n = 6 k + 4 = 2 + 3 + \paren {6 k - 5}$

All terms are greater than $1$.

The first two terms are powers of $2$ and $3$, so they are coprime.

The last term is not divisible by $2$ or $3$, so the three numbers are pairwise coprime.

Case $2$: $n$ is odd
There is some integer $k \ge 1$ such that one of the following holds:


 * $n = 12 k + 1 = 3 + \paren {6 k - 7} + \paren {6 k + 5}$ and $k \ge 2$


 * $n = 12 k + 3 = 9 + \paren {6 k - 5} + \paren {6 k - 1}$ and $k \ge 2$


 * $n = 12 k + 5 = 3 + \paren {6 k - 5} + \paren {6 k + 7}$ and $k \ge 2$


 * $n = 12 k + 7 = 3 + \paren {6 k - 1} + \paren {6 k + 5}$


 * $n = 12 k + 9 = 3 + \paren {6 k - 1} + \paren {6 k + 7}$


 * $n = 12 k + 11 = 3 + \paren {6 k + 1} + \paren {6 k + 7}$

All terms are greater than $1$.

The first term is a power of $3$.

The last two terms are not divisible by $3$, so they are coprime with $3$.

the last two terms are not coprime.

Then they must have a common divisor greater than $1$.

This common divisor must also divide their difference.

However their difference is only divisible by $2$ and $3$, and the terms themselves are not.

Hence they cannot have a common divisor greater than $1$.

So they must be coprime.

Hence the result.