Does there exist a convex $n$-gon with all sides equal and all vertices on the graph of $y=x^2$ Does there exist a convex $n$-gon with all sides equal with all vertices lying on graph $y=x^2$ when


*

*$n$ is odd

*$n$ is even
I think triangle may be right, but I don't know how to prove that

 A: Here are two particular solutions: (1) for the triangle (I have repeated here what I have given as a comment before), (2) for a pentagon. In this last case, we reach a limit of algebraic calculation.
(1) For the case of the triangle you have given : let  $A(a,a^2)$ and  $A'(a,a^2)$. Expressing the equality of the squares of distances $AA'^2=OA'^2$ gives $(2a)^2=a^2+a^4$. Thus $3a^2=1 \Rightarrow a=1/\sqrt{3}.$
(2) or the case of a pentagon with vertices:
$$O(0,0), \ \ A(a,a^2), \ \ B(b,b^2), \ \ B'(-b,b^2), \ \ A'(-a,a^2)$$     
we have 2 constraints   $OA^2=AB^2=BB'^2$ yielding two equations:
$$a^2+a^4=(a-b)^2+(a^2-b^2)^2=4b^2 \ \ \ \ (1)$$
By elimination of $b$ between equations (1), one gets:
$$(80 + 40A + 13A^2 - 10A^3 + A^4)(A-3)=0 \ \ \ (2) \ \ \ \text{where} \ \ \ A:=a^2  $$
(knowing $A$, one immediately gets $b=\sqrt{A+A^2}/2$, using (1)).
Let us first consider the simple solution $A=3 \Rightarrow a = \sqrt{3} \Rightarrow b=\pm \sqrt{3}$ (using (1)): these values of $b$ are clearly non valid. Thus this does not lead to a solution.
The other exact solutions of (2) cannot be tackled in an exact manner. We have to turn to approximate solutions, which are:


*

*$A = 7.2190025 \Rightarrow a = 2.6868 \Rightarrow b=3.8514$ which are convenient. But there is another real root :

*$A= 4.7577995  \Rightarrow a = 2.18124 \Rightarrow b=-2.61699$ which correspond to a self-intersecting polygon, which, evidently, is not convex.

*the two other roots of (2) are complex.
Consequence : this problem has a solution for the triangular and the pentagonal cases.
Edit: I add the quadrilateral case: As a quadrilateral with equal sides' lengthes is a lozenge, we are going to see that no lozenge can be inscribed in a parabola.
Let us assume that a lozenge $ABCD$ (described in direct orientation) is such that $A(a,a^2), B(b,b^2), C(c,c^2), D(d,d^2)$. Then, the midpoints of the diagonals should coincide, i.e., 
$$(\dfrac{a+c}{2},\dfrac{a^2+c^2}{2})=(\dfrac{b+d}{2},\dfrac{b^2+d^2}{2})$$
But a little computation show that it would imply: 
$$\begin{cases}a+c&=&b+d\\
ac&=&bd\end{cases}$$
But there is a unique pair of numbers having a given sum and a given product.
Thus either $a=b$ and $c=d$ or $a=c$ and $b=d$ : in both cases, it's a degenerate case. 
