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This question is obvious in the sense that prime gaps on average are larger as the numbers go towards infinity.

However, I don't have any idea on how to begin tackling this question apart from very basic trial and error.

Any ideas will be useful. Thanks.

Robert Soupe
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Haikal Yeo
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  • Even plus even equals odd. 9 is odd. So any odd prime equal or large than 17 can be written as composite even plus composite even +9. – fleablood Jun 05 '17 at 03:09

3 Answers3

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Look at the sequence of composite numbers: 4, 6, 8, 9, 10, 12, 14, 15, 16, 18, etc. Just try choosing three of them at random, adding them up and taking note of what happens.

Do the composite numbers have to be distinct? If so, the smallest prime that can be expressed as the sum of three distinct composite numbers is 4 + 6 + 9 = 19. Then, for any $p > 17$, we can just subtract 9 and the problem becomes one of finding two distinct even composite numbers to add up to $p - 9$. The easiest way, I think, is $$\left(\frac{p - 9}{2} - 1\right) + \left(\frac{p - 9}{2} + 1\right).$$

If you don't require the composite numbers to be distinct, then you can also do 9 + 4 + 4 = 17.

Robert Soupe
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    That is guaranteed to always work if $p \equiv 3 \pmod 4$. But if $p \equiv 1 \pmod 4$ it might not always work (e.g., $p = 29$, but then you can use $\pm 2$ instead of $\pm 1$. A minor detail, but like they say, the demon is in the details! Mwahahahahahahahaha! – The Short One Jun 05 '17 at 20:59
  • Good point, thanks. – Robert Soupe Jun 06 '17 at 01:42
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Hint: any prime $\ge 17$ can be written as $9 + 4 + y$ where $y$ is a composite number.

Robert Israel
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Hint: All odd number greater than or equal to $17$ can be written as a sum of three composites: $9 + 4 + (4+2j)$ for some $j\ge 0$.

paw88789
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