n+2=a^3Originally Posted byCaptainBlank

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.

Results 1 to 4 of 4

- Feb 5th 2007, 06:00 AM #1

- Joined
- Nov 2005
- From
- someplace
- Posts
- 14,972
- Thanks
- 5

- Feb 11th 2007, 10:05 AM #2

- Joined
- Nov 2005
- From
- New York City
- Posts
- 10,616
- Thanks
- 10

- Feb 12th 2007, 09:14 AM #3

- Joined
- Nov 2005
- From
- someplace
- Posts
- 14,972
- Thanks
- 5

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.

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

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 #4

- Joined
- Nov 2005
- From
- New York City
- Posts
- 10,616
- Thanks
- 10