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Integer solutions

  1. Dec 11, 2007 #1
    how does one go about finding integer solutions for an equation such as this? Is it easier to merely find how many solutions?

    y^2 = x^3 + n, n is some integer.
     
  2. jcsd
  3. Dec 12, 2007 #2
    if n = +- 2 this is a classic Fermat's problem: prove that 26 is the only natural number between an square and a cube (5^2 = 25 and 3^3 = 27)
     
  4. Dec 12, 2007 #3
    what if n is large? say n>100
     
  5. Dec 12, 2007 #4
    where did you find this problem? Could you put here with the exacly words?
     
  6. Dec 13, 2007 #5
    I thought X^2=Y^3 + 1 was a well-known problem.

    There are trivial solutions, such as y=0, x=1, or x=0, y=-1. Ingnoring that, we have (X-1)(X+1) = y^3, with all factors assumed positive. X-1 and X+1 can only have 2 as a common factor. And, in that case, one term is divisible by 4. X-1=2a^3, X+1 = 4b^3, assuming a>0, b>0, this gives us 2=4a^3-2b^3, or 1 =2a^3-b^3. This is only possible if a=b=1. Thus the solution is 3^2=2^3+1.

    For negative terms,b=-1, a=-1, 1=b^3-2a^3, giving x-1=4a^3=-4, x+1=2b^3=-2, we could have found, but excluded: (-3)^2= 2^3+1 for a=b=-1.

    If they have nothing in common we get 2 = b^3-a^3, which gives no answer in positive integers. However, X-1= a^3=-1, X+1= b^3=1, gives X=0, Y=-1, another trivial result.

    Thus there is only one non-trivial solution, already given: 3^2= 2^3+1.
     
    Last edited: Dec 13, 2007
  7. Dec 13, 2007 #6
    That is the solution where n = 1, but more importantly what the poster asks is what method can be used to obtain the solution if n = other, e.g. 22. One could pose an infinite number of such questions but I doubt that there is a direct methof of solution. At least there is no discussion of such a problem as far as I know.
     
    Last edited: Dec 13, 2007
  8. Dec 13, 2007 #7
    there are infinite n such that n = x^2 - y^3, this is trivial

    there are at least one x and one y such that n = x^2 - y^3 for every n integer? this is the question, I think... it's hard!
     
  9. Dec 13, 2007 #8
    ramsey2879: At least there is no discussion of such a problem as far as I know.

    I agree. I don't think anything has been done beyond piecemeal efforts. Obviously, n=1 is a simple case.

    al-mahed: if n = +- 2 this is a classic Fermat's problem: prove that 26 is the only natural number between an square and a cube (5^2 = 25 and 3^3 = 27)

    I have some memory of 3^3=5^2+2. I can't seem to find any information about these problems at all.
     
    Last edited: Dec 13, 2007
  10. Dec 13, 2007 #9
    al-mahed: there are at least one x and one y such that n = x^2 - y^3 for every n integer? this is the question, I think... it's hard!
    That, I feel is not true. Take 4=X^2-Y^3. Obviously Y=0 is such a solution, a negative Y impossible. But for X and Y greater than 0, consider them both even Then in this case we have 4=4x'^2+8y'^2. This leaves us 1=x'^2+2y'^3. Only solution y' = 0.

    Now if X and Y are both odd,Y^3=(X-2)(X+2), assume X>2, these have no common factor, (since only 2 could be such a factor) x-2 =a^3, x+2 = b^3. This then gives
    4 = b^3-a^3 = (b-a)(b^2+ab+a^2). Since a and b are odd, 4 divides b-a, and thus we have the form: [tex]1= (b^2+ab+a^2)\frac{b-a}{4}[/tex] Clearly b-a can not be zero and with only positive terms a^2+ab+b^2 exceeds 1.
     
    Last edited: Dec 14, 2007
  11. Dec 13, 2007 #10
  12. Dec 17, 2007 #11
    I did not follow you here: Now if X and Y are both odd,Y^3=(X-2)(X+2), assume X>2, these have no common factor, (since only 2 could be such a factor) x-2 =a^3, x+2 = b^3.

    why did you conclude that x-2 and x+2 can each be written as a cube?
     
  13. Dec 17, 2007 #12

    CRGreathouse

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    He took n = 4, which means 4 = X^2-Y^3. Adding Y^3 - 4 to each side gives Y^3 = X^2 - 4 = (X - 2)(X + 2). Since they have no common factors, if one is not a cube then the product is not a cube.
     
    Last edited: Dec 17, 2007
  14. Dec 17, 2007 #13
    There are no cubes which are 4 apart. The difference between consecutive cubes grows rapidly (quadratically).
     
  15. Dec 17, 2007 #14
    Now I see, thanks!

    I made my question because I had conclude that if only one is a cube the product can not be a cube, but is you right when you say: both need to be a cube, and if no one is a cube, there is no way an odd number N with the same factor than N + 4
     
  16. Dec 17, 2007 #15
    Yes, this part is clear, I made another question, concerning the fact that y^3=(x-2)(x+2) ==> y| (x-2) or (x+2) only if y is prime (of course y dividing only one makes impossible the product be a cube). If y is not prime was not so clearly for me that (x-2)(x+2) cannot form a cube, although to be almost obvious.
     
  17. Dec 24, 2007 #16
    What ever prime divides (x-2) and (x+2) also divides the sum, which is 4.

    The fundamental theorem of arithmetic tells us that any integer can be decomposed into prime factors in only one way, without regard for order. Thus we only need to consider prime factors.
     
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