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 $n$?
Solution 1:
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.
Solution 2:
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.