Two easy ones this week:

1.Let be a positive integer. Prove that the numbers and cannot both be perfect cubes.

2. Which regular n-gons can be inscribed in a non-circular ellipse?

RonL

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Truth does not change because it is, or is not, believed by a majority of the people.

Giordano Bruno

n+2=a^3
n^2+n+1=b^3
Thus,
(n^2+n+1)(n+2)=a^3b^3=(ab)^3=m^3
(n^2+n+1)((n-1)+3)=m^3
(n^2+n+1)(n-1)+3(n^2+n+1)=m^3
n^3-1+3n^2+3n+3=m^3
(n^3+3n^2+3n+1)+1=m^3
(n+1)^3+1^3=m^3
Fermat's Last Theorem n=3.

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"Democracy has proved only that the best way to gain power
over people is to assure the people that they are ruling
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If both n+2 and n^2+n+1 are both cubes then so is their product, but:

(n+2)(n^2+n+1)=(n+1)^3+1

but this is imposible as no two cubes of integers differ by 1.

A regular n-gon can be inscribed in a circle, but the vetices also lie on
the non-circular ellipse. But a pair of distinct conics intersect at no more
than four points, so n<=4.

RonL

__________________
Truth does not change because it is, or is not, believed by a majority of the people.

Giordano Bruno

For some reason it seemed to me you where asking for which regular n-gons are constructable on a non-circular ellipse with a compass and straightedge.

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"Democracy has proved only that the best way to gain power
over people is to assure the people that they are ruling
themselves. Once they believe that, they make wonderfully
submissive slaves." - Joseph Sobran