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Diophantine equations

  1. Feb 5, 2007 #1
    i want to find solutions to the equations:
    1. [tex]\left(3^x - 2^y\right)^2 = 1[/tex] (x and y are integers)
    2. [tex]a^3 - 1 = b^2[/tex] (solutions should be positive integers)

    i can "see" that two solutions of the first equation are (1, 1) and (2, 3)
    but how can i find the other solutions?

    i have seen an equation similar to the second one: [itex]a^3 - 2 = b^2[/itex] which have only 2 solutions (3, 5) and (3, -5). but how do i solve the equation where the constant is -1? is there a general way to solve equations of the form [itex]a^3 + m = b^2[/itex]?

    as i am self studying, these equations seem really complicated for me to solve, although experts on number theory might find them trivial. i apologize to them for asking such trivial (to them) questions.
    Last edited: Feb 5, 2007
  2. jcsd
  3. Feb 5, 2007 #2
    Catalan's theorem says that (1,1) and (2,3) are the only solutions to the first one, but I do not know if there is an elementary way of seeing this.

    As for the second problem, you can rewrite it as [tex]a^3=b^2+1[/tex], then factor over [tex]\mathbb{Z}[/tex] to get [tex]a^3=(b+i)(b-i)[/tex]. You can check that [tex]b+i[/tex] and [tex]b-i[/tex] are relatively prime, thus each must be a perfect cube. Thus you would have
    [tex](\alpha +i\beta)^3=\alpha^3 +3i\alpha^2 \beta -3\alpha \beta^2 -i\beta^3=b+i[/tex], where [tex]\alpha,\beta[/tex] are integers. Equating real and imaginary parts, you get
    [tex]\alpha^3-3\alpha\beta^2 = b [/tex] and
    [tex]3\alpha^2\beta -\beta^3=1 [/tex].
    Factoring the last equation gives
    So [tex]\beta=\pm 1 [/tex], and you can use this to figure out [tex]\alpha [/tex], since [tex]3\alpha^2-1=\mp 1 [/tex]. I think it gives [tex]\alpha=0 [/tex] as the only possible solution. Thus b=0, a=1 is the only possible solution.

    I believe this is a common strategy for solving such types of Diophantine equations. That is to say, factoring over a larger number ring. The only problem is that unique prime factorization doesn't hold in all of these rings, so then you pass to ideals, and then there are some issues with the class number.

    These are by no means trivial questions, and a lot of awesome math goes into solving problems like these.
    Last edited: Feb 5, 2007
  4. Feb 6, 2007 #3
    what is the Catalan's theorem?
    how can i show that b+i and b-i are relatively prime?
    and how did you get the following:

  5. Feb 6, 2007 #4
    Any prime dividing a+bi and a-bi has to divide the sum =2a and the difference, 2 bi, so that only 2 need be considered, which splits into 1+i and 1-i. (2 is an unusual case since (1+i)/(1-i)=2/(-2i) =i, so they differ only by a unit.)

    Take the case of (a+bi)/(1+i), multiply by (1-i)/(1-i) = (a-b + (a+b)i)/2, which means both terms have the same parity, if it is divisible. Primes, however, other than 2 are odd, so that primes that split have factors such as a+bi, with one term even and other odd.

    In the case given, a^3=b^2+1, we know that a is odd. Why? because if b is odd, an odd square is congruent to 1 Mod 8, so that b^2+1 is divisible by no more than 2, but if a were even, a^3 would be divisible by 8.

    I see that I only addressed your second question. Since there is a lot of material involved, for question 3 you should look up: Gaussian Integers. For question 1, http://primes.utm.edu/glossary/page.php?sort=CatalansProblem
    Last edited: Feb 7, 2007
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