The cube sequence is,
1,8,27,64,125,....
The triangular sequence is,
1,3,6,10,15,21,...

Show that besides for 1. There is no other number that is simulatenously a cube and a triangle.

(HINT: Use the Catalan Conjecture, which was proven in 2002. Now called Mihăilescu's theorem).

__________________

To view links or images in signatures your post count must be 10 or greater. You currently have 0 posts.


"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

Absolutely right, TPHacker! . . . *blush*

Everybody, please ignore this egregious incorrect solution . . .

The cubes are: n³
The triangular numbers are: n(n + 1)/2

Then we have: n³ = n(n+1)/2 ---> 2n³ - n² - n = 0

Factor: n(n - 1)(2n + 1) = 0

Since n is a positive integer, the only root is: n = 1

Because the n-th triangle is not necessarily the n-th cube. You can have, say, the 20-th triangle be equal to the 11-th cuve, note they are not the same.

__________________

To view links or images in signatures your post count must be 10 or greater. You currently have 0 posts.


"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

This is my problem, but the solution is not. As I said it relys on the Catalan conjecture. A specific type of Catalan equation. The first ever considered was:
x² = y³ +1
It was shown, by Euler I believe, that the only solutions is,
x=3 and y=2.
If you want to know why, I cannot tell you, all books on Number theory say that is proof exists but it is far too long to post.
However, there is a simpler way to demonstrate this, if you are familar with elliptic curves (I am not) it is a related to Mordell's theorem,
x² = y³ + k.
The cases k=2,4 where discussed by Fermat and are even more complicated then k=1 (this case). Which thats they are at most a finite number of solutions.

Now we can return to the problem. The solution was given to me by a math professor I like to discuss math with.

We need to show,
x(x+1)/2 = y³
Has no solutions.
Multiply by 8,
4x(x+1)=8y³
4x²+4x=8y³
Add one,
4x²+4x+1=8y³+1
Complete the square and cube,
(2x+1)²=(2y)³+1
By Fermat/Euler/Mordell/Catalan or whatever we have,
2x+1=3
2y=2
Thus,
x=1
y=1
Are the only solutions.
That means only the first terms in these sequences are simulanteous cubes and triangles.

__________________

To view links or images in signatures your post count must be 10 or greater. You currently have 0 posts.


"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