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I find this beautiful

  1. Feb 13, 2012 #1
    I was studying (yet another) number theory problem, described here:

    Prove that the equation x^3+y^3+z^3+t^3=1999 has infinitely many solutions (x,y,z,t) in integers.

    I found a way of constructing these solutions, which I will describe right now: Consider the quadruple of real numbers of the form (10+n,10-n,-(60n^2)^1/3,-1), where n is an integer. The sum of the cubes of these real numbers, for arbitrary n, is always 1999. Let n be of the form s^3*60, where s is any integer. Then the quadruple described above becomes an integer quadruple. Since there are infinitely many such n, there are infinitely many integer solutions to the equation.

    I found this fascinating! Can you all find an alternate way of solving the problem? I'd be interested.

    mathguy15(whos now 16)
  2. jcsd
  3. Feb 13, 2012 #2
    Hi Mathguy, nice work. I tried for more than reasonable time without getting there. I strongly believe you have *the* solution, and nothing substantially different will come up.
  4. Feb 14, 2012 #3
    Thanks. I wonder if something related to group theory can be applied. Sadly, I dont know enough group theory to make it work :(. But that's ok.
  5. Feb 15, 2012 #4
    Oh, and I have found a generalization to all integers of the form (2x^3)-1. In the quadruple of real numbers, replace 10 by x and and -(60n^2)^1/3 by -(6xn^2)^1/3, and then n is just of the form (s^3)*6x for some integer s. This produces a way of representing any number of the form (2x^3)-1 as the sum of four integer cubes.

    EDIT:and there is a method for representing cubes as the sum of four cubes and integers that are twice some cube as the sum of three cubes.
  6. Feb 15, 2012 #5
    Binomial expansion.



  7. Feb 15, 2012 #6
    You can probably generalize further, and replace the 1 in 2x3-1 with any cube.
  8. Feb 16, 2012 #7
    Yeah! For any cube(r^3) and any integer(n), the quadruple of real numbers of the form {n,-n,0,r} satisfies the equation x^3+y^3+z^3+t^3=r^3.

    EDIT:Sorry, I thought you meant replace the 2x^3-1 by any cube. In order to generalize to the case you mentioned above, in the quadruple of real numbers {10+n,10-n,-(60n^2)^1/3,-1}, just replace the -1 by any integer, and you get infinitely many representations of the number 2x^3-f^3 as the sum of four cubes.
    Last edited: Feb 16, 2012
  9. Feb 16, 2012 #8
    The further problem here is to find (and to prove that you found) all the solutions.

    You found infinitely many solutions, which is pretty good. Now try to find them all (if there are any).
  10. Feb 16, 2012 #9
    So the next step is to classify every type of solution that could arise?
  11. Feb 16, 2012 #10
    That would indeed be an interesting problem.
  12. Feb 16, 2012 #11
    Sorry, I'm new at this. Could you give this newbie some inkling of what that problem could entail?
  13. Feb 16, 2012 #12
    The problem is likely very difficult. It is a problem in Diophantine equations, which can turn out to be quite complicated. Furthermore, your problem has 4 variables, which doesn't help.

    Even the easier problems of elliptic curves are very difficult.
  14. Feb 16, 2012 #13
    Hey! Those were used in the proof of Fermat's last theorem. I don't really know anything about elliptic curves, but I want to be a mathematician, and I like number theory, so I might be involved with those in about 6-7 years.
  15. Feb 16, 2012 #14
    It's a beautiful theory. Check out the website http://mathcircle.berkeley.edu/BMC4/Handouts/elliptic/node1.html [Broken] to see what kind of methods are involved. With a knowledge of geometry, you should be able to understand it.
    Last edited by a moderator: May 5, 2017
  16. Feb 16, 2012 #15


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    I'm gonna go ahead and guess that the problem of determining all integer solutions to
    [tex]x^3+y^3+z^3+t^3=1999[/tex]is open.

    I have two reasons to suspect this: 1) These kinds of problems are generally really, really, really difficult to solve (cf. FLT, Erdos-Strauss, ...). 2) Let's suppose we were able to classify all integer solutions. We would then, in particular, be able to classify integer solutions in which t=1, or equivalently, we'd be able to solve
    [tex]x^3+y^3+z^3=1998=3^3\cdot 74[/tex]in the integers. If we could solve that, then we'd know a thing or two about the rational solutions to
    [tex]X^3 + Y^3 + Z^3 = 74.[/tex] I believe nobody knows if this last equation has any integer solutions. Indeed, in Cohen, Number Theory: Analytic and Modern Tools (Springer, 2007), one finds the following piece of info:
    This leads to me suspect that people don't know much about the rational solutions of this equation either (but I could be wrong).
    Last edited: Feb 16, 2012
  17. Feb 16, 2012 #16
    I don't follow. If one classifies the solutions with t=-1, wouldn't that mean one could solve x^3+y^3+z^3=2000 in integers rather than 1998?
  18. Feb 16, 2012 #17


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    Yes, sorry - that was a typo. I meant t=1.
  19. Feb 16, 2012 #18
    It would be interesting if one could prove that 1 is never one of the integers in the solution.

    EDIT:Oh, and thank you micromass for that link. I will definitely look into it.
  20. Feb 16, 2012 #19
    It will be since

  21. Feb 16, 2012 #20
    Wow, how did you get that? Did you use some kind of computer or did you find it using some theory or theorem?
  22. Feb 16, 2012 #21
    Just letting a computer check the possibilities. Nothing fancy.
    I didn't even program it, I found this on a list online :tongue2:
  23. Feb 16, 2012 #22
    Hi Mathguy15,

    Multiply out the terms, but first cube your 3rd term since you're going to cube it later anyway...

    n^3 + 30*n^2 + 300n + 1000
    -n^3 + 30*n^2 - 300n + 1000
    - 1

    30*n^2 + 30*n^2 - 60*n^2 = 0
    1000 + 1000 - 1 = 1999

    You could do the same thing for (n+8)^3 = 1n^3 + 24*n^2 + 192n + 512, and (n-8)^3, but you'd have to replace (-60*n^2) with (-48*n^2) and 1999 with 1023. And so on...

    Note that you're not really dealing with cubes since they cancel out when you multiply out and collect the terms.
  24. Feb 17, 2012 #23
    Seems you're right!

  25. Feb 17, 2012 #24
    If you want to generalize further mathguy15...

    for any n or k in N

    First term: (1*k^0*n^3) + (3*k^1*n^2) + (3*k^2*n^1) + (1*k^3*n^0) = (n+k)^3
    Second term: -(1*k^0*n^3) + (3*k^1*n^2) - (3*k^2*n^1) + (1*k^3*n^0)
    Third term: (Derived from above 2nd terms doubled)
    Fourth term: -1

    ... then figure out the formula for those coefficients (x) above (which is almost certainly related to the Binomial Coefficients aka "the coefficients of the expansion of (x+1)^n"). Do that and then you'll have a nice little equation in 3 variables that will work for any n, k or x in N.

    Binomial Coefficients:
    1 1 --> Powers of 1
    1 2 1 --> Powers of 2
    1 3 3 1 --> Powers of 3...
    1 4 6 4 1 --> Powers of 4...

    ...when inserted into Polynomials.

    In the example you posted, then set k = 10

    - AC
    Last edited: Feb 17, 2012
  26. Feb 17, 2012 #25
    Here's an example of what I was referring to in the previous post, thinking-wise. Take the form...

    (10 + n)^5 + (10 - n)^5 + (-100*n^4) + (-20000*n^2) + (-1) = 199999

    Echoing the form used by mathguy15...

    v = (10 + n)
    w = (10 - n)
    x = (-100*n^4)^(1/5)
    y = (-20000*n'^2))^(1/5)
    z = -1

    x = an integer for (-100*(100*s^5)^4)^(1/5); n =(100*s^5)
    y = an integer for (-20000*(125*s'^10)^2))^(1/5); n' = (125*s^10)

    Will have to see where to go from there, meaning what conclusions one might come to regarding:
    v^5 + w^5 + x^5 + y^5 + z^5
    Last edited: Feb 17, 2012
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