# Problem 18

• Feb 5th 2007, 06:00 AM
CaptainBlack
Problem 18
Two easy ones this week:

1.Let \$\displaystyle n\$ be a positive integer. Prove that the numbers \$\displaystyle n+2\$ and \$\displaystyle n^2+n+1\$ cannot both be perfect cubes.

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

RonL
• Feb 11th 2007, 10:05 AM
ThePerfectHacker
Quote:

Originally Posted by CaptainBlank
1.Let \$\displaystyle n\$ be a positive integer. Prove that the numbers \$\displaystyle n+2\$ and \$\displaystyle n^2+n+1\$ cannot both be perfect cubes.

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.
• Feb 12th 2007, 09:14 AM
CaptainBlack
Quote:

Originally Posted by CaptainBlack
Two easy ones this week:

1.Let \$\displaystyle n\$ be a positive integer. Prove that the numbers \$\displaystyle n+2\$ and \$\displaystyle n^2+n+1\$ cannot both be perfect cubes.

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.

Quote:

2. Which regular n-gons can be inscribed in a non-circular ellipse?
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
• Feb 12th 2007, 10:36 AM
ThePerfectHacker
Quote:

Originally Posted by CaptainBlack
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.