Conjecture: Every $n \geq 20 \in \mathbb{N}$ can be written as a sum of three integers $(\geq 2)$ that are pairwise coprime

Question

This question on the sum of pairwise prime numbers piqued my interest, and I started looking at what numbers can be written as the sum of three pairwise coprime numbers (excluding $1$): $$ \begin{align*} 10&=5+3+2\\ 12&=7+3+2\\ 14&=7+5+2\\ 15&=7+5+3\\ 16&=9+5+2\\ 18&=11+4+3\\ 19&=11+5+3\\ 20&=11+7+2\\ 21&=11+7+3\\ 22&=11+7+4\\ &\ldots \end{align*} $$ It's fairly clear why $10$ is the smallest number possible, but I couldn't find any pattern until $20$ where it seems like all the rest follow suit.

It seems after $20$, there are so many possible combinations that every following number can be written as a sum of three integers $(\geq 2)$ which are pairwise coprime. Is this a known result, and, can it be (dis)proved?

EDIT: Thanks for the responses all, I believe I have devised a nice case proof (inspired by Yuval Peres' case answer!)

Proof

Case 1: $n$ is even $$ \begin{align*} n&=6k=2+3+(6k−5)\\ n&=6k+2=4+3+(6k−7)\\ n&=6k+4=2+3+(6k−5) \end{align*} $$ Case 2: $n$ is odd $$ \begin{align*} n&=12k+1=3+(6k−7)+(6k+5) \text{ with } k≥2\\ n&=12k+3=9+(6k−5)+(6k−1) \text{ with } k≥2\\ n&=12k+5=3+(6k−5)+(6k+7) \text{ with } k≥2\\ n&=12k+7=3+(6k−1)+(6k+5)\\ n&=12k+9=3+(6k−1)+(6k+7)\\ n&=12k+11=3+(6k+1)+(6k+7) \hspace{35pt}\blacksquare \end{align*} $$

Answer 1

If $n \ne 2 \!\!\! \mod 3$ and $n \ne 3 \!\!\! \mod 5$: $\;\;$ Write $\;n=3+5+(n-8)$ provided $n \ge 10$.

If $n \ne 2 \!\!\! \mod 3$ and $n = 3 \!\!\! \mod 5$: $\;\;$ Write $\;n=9+5+(n-14)$ provided $n \ge 16$.

If $n =2 \!\!\! \mod 3$ and $n \ne 3 \!\!\! \mod 5$: $\;\;$ Write $\;n=3+25+(n-28)$ provided $n \ge 30$. $\quad$ Observe that 26=2+5+19.

If $n =2 \!\!\! \mod 3$ and $n = 3 \!\!\! \mod 5$: $\;\;$ Write $\;n=9+25+(n-34)$ provided $n \ge 36$. $\quad$ Observe that 23=3+7+13.

Answer 2

Assume that $n$ is not the power of sum prime, i.e $n$ cannot be written as $n=p^{\alpha}$ for some prime $p$ and exponent $\alpha$. In this case, we can find two distinct primes $p,q$ that both divide $n$. Clearly $p+q<pq<n$ and so $n-(p+q)$ is a positive integer. This means that we can write

$$n=p+q+(n-(p+q)).$$

The first two terms, $p$ and $q$, are clearly coprime since $p$ and $q$ are prime, and hence all there is left to check is that $(n-(p+q))$ is not a multiple of $p$ or $q$. This follows from the fact that

$$n-(p+q)\equiv -q \neq 0\mod(p)$$

$$n-(p+q)\equiv -p \neq 0\mod(q)$$

since $n$ is a multiple of $p$ and $q$, and both $p$ and $q$ are prime thus are nonzero modulo each other.

In the general case you should be able to do something similar by shifting, since prime powers are 'rare enough' in a certain sense.