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

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.

__________________
And he (Elisha) went up from thence unto Bethel: and as he was going up by the way, there came forth little children out of the city, and mocked him, and said unto him, "Go up, thou bald head"; "go up, thou bald head". And he turned back, and looked on them, and cursed them in the name of the Lord. And there came forth two she-bears out of the wood, and tore up forty and two children of them.
Second Kings 2: 23-24

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

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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And he (Elisha) went up from thence unto Bethel: and as he was going up by the way, there came forth little children out of the city, and mocked him, and said unto him, "Go up, thou bald head"; "go up, thou bald head". And he turned back, and looked on them, and cursed them in the name of the Lord. And there came forth two she-bears out of the wood, and tore up forty and two children of them.
Second Kings 2: 23-24