# Do 4 consecutive primes always form a polygon?

Related to this question, if 4 segments have length of 4 consecutive primes, can they always form a 4-vertex polygon?

This question occurred to me out of sheer curiosity, but now I can't prove or disprove it, and I can't sleep knowing that.

According to one form of Bertrand's postulate, $p_ {n+1} < 1.1 \times p_{n}$ for large enough $n$, so it is easy to prove that for large enough $n$, the statement about polygon is true. But how to know the value of "large enough $n$", so that the statement about polygon can be manually checked for smaller $n$?

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Would it help to know the result of Erdős, that for any sufficiently large $n$, there are four primes between $n$ and $2n$? – hardmath Jul 10 '14 at 21:02
Some specifics of "large enough $n$" are given here with link to OEIS. – hardmath Jul 10 '14 at 21:05

Per the question you linked, the largest $3$ out of any $4$ consecutive primes form a triangle. Keeping one side fixed in place, disconnect the opposite corner and swing the other two sides outward (with each side rotating around the corner where it touches the fixed side), until all three sides form a single line segment.

During this process, the distance between the moving endpoints starts at $0$ and continuously changes until it is the sum of the larger $3$ primes, so by the Intermediate Value Theorem at some point it is equal to the smallest prime. Connect the endpoints and you've got your quadrilateral.

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Are you also saying that similar statement with N consecutive primes can be proven by induction? – VividD Jul 10 '14 at 21:20
Yes, you could do a similar trick with any $(N-1)$-gon, though it would be a little more involved to describe how to "unfold" the edges into a straight line in a rigorous way. – MartianInvader Jul 10 '14 at 22:53

Yes.

To have the lengths be valid for a quadrilateral, any one side must have length less than the sum of the lengths of the other three sides.

Bertrand's Postulate ensures that this is easily satisfied for four consecutive primes.

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This is a nicely observed Answer. Only the largest of any $k=4$ consecutive primes needs to be checked (as the lesser ones are clearly bounded by the largest one) and the same can be said for $k>4$. – hardmath Jul 11 '14 at 12:10