The cube sequence is,
The triangular sequence is,
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).
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.
Originally Posted by Soroban
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,
Complete the square and cube,
By Fermat/Euler/Mordell/Catalan or whatever we have,
Are the only solutions.
That means only the first terms in these sequences are simulanteous cubes and triangles.