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This problem is pretty hard :

Does there exist a convex $n$-gon with all sides equal, whose vertices lie on the graph $y=x^2$ when $n$ is an even number ?

The case of $n$ being odd can be delt with after some manipulations (the answer is yes). The even case seems more tricky.

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The convexity condition forces the vertices to be aligned on the graph one after another passing from the highest vertex on the right to the highest vertex on the left.

If the origin is not a vertex, then it is obvious that the distance of the two highest vertices is larger than the distance of the two lowest vertices.

So, the origin is a vertex and the layout of the vertices on the two branches is symmetrical except for one additional vertex, say on the right (because of even parity).

The highest vertex $L$ on the left, and the two highest vertices on the right $R_1$ (highest) and $R_2$ form a triangle with $\measuredangle LR_2R_1$ being larger then 90 degrees. But you also know that the distances $R_1R_2$ and $R_1L$ are the same which gives a contradiction.

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"The highest vertex L on the left, and the two highest vertices on the right R 1 (highest) and R 2 form a triangle with ∡LR 2 R 1 being larger then 90 degrees. " Could you elaborate this part ? Thanks –  momo1729 Apr 15 '12 at 11:33
    
$L$ and $R_2$ are at the same height, $R_1$ is higher and to the right of $R_2$, so the angle is larger than 90 degrees. –  Phira Apr 20 '12 at 11:31
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