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

Clyde Kertzer
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  • What about 2, 5 and 7; or 3, 5, and 7? – Joe Apr 24 '22 at 00:30
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    Are you familiar with Goldbach's conjecture? – Milo Moses Apr 24 '22 at 01:49
  • @MiloMoses, yes I am. – Clyde Kertzer Apr 24 '22 at 01:54
  • Goldbach's conjecture implies this for $n$ even as you can express $n-2$ as the sum of two odd primes and add $2$. For odd $n$ you have the conjecture, which is easier but as far as I can tell unproven, that you can express it as the sum of three primes. – Ross Millikan Apr 24 '22 at 02:57
  • @RossMillikan odd/weak/ternary Goldbach (every odd integer $>5$ can be expressed as the sum of 3 primes) is proven; for sufficiently large odd $n$ by Vinogradov in 1937, and for all odd $n>5$ by Helfgott recently: https://math.stackexchange.com/questions/3164020/the-significance-and-acceptance-of-helfgott-s-proof-of-the-weak-goldbach-conject – D.R. Nov 10 '22 at 07:14

2 Answers2

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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.

Yuval Peres
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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.

Milo Moses
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