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Math Help - Question 9

  1. #1
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    Question 9

    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).
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  2. #2
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    Why Catalan?

    Absolutely right, TPHacker! . . . *blush*

    Everybody, please ignore this egregious incorrect solution . . .


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

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

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

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

    Last edited by Soroban; December 5th 2006 at 11:08 AM.
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  3. #3
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    Quote Originally Posted by Soroban View Post

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

    Then we have: n = n(n+1)/2 ---> 2n - n - 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.
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  4. #4
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    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:
    x = y +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,
    x = y + 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 = y
    Has no solutions.
    Multiply by 8,
    4x(x+1)=8y
    4x+4x=8y
    Add one,
    4x+4x+1=8y+1
    Complete the square and cube,
    (2x+1)=(2y)+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.
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