Show that there are no knotted quadrilaterals or pentagons. what knot types are represented by hexagons? by septagons? I am reading the book of Richard H. Crowell of "Introduction to knot theory" And in chapter I, it has this question Show that there are no knotted quadrilaterals or pentagons. what knot types are represented by hexagons? by septagons? And I found this question on this side How to prove a regular pentagon is formed by knotting a rectangular strip of paper? Are not the two questions contradicting each other? How can I answer the first question in a simple way?

Think about how many crossings you can get if you have a quadrilateral. I would say that you can have at most one crossing, but that means your knot is trivial and therefore there are no knotted quadrilaterals.

Now a similar argument works for pentagons.

I do not think that there are only unknotted hexagons as you said since the trefoil knot is for example given by a hexagon. Actually, that should be the only non-trivial one since the trefoil knot and its mirror image are equivalent by your definition of equivalence.

For heptagons the only non-trivial knot that arises should be the unique knot with crossing number $4$, the figure eight knot.

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