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Math Help - Gaussian integers

  1. #1
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    Gaussian integers

    Use the Gaussian integers to find the solutions in rational integers of the diophantine equation x^2+1=y^3

    a) Show that if x and y are integers such that x^2+1=y^3, then x - i and x + i are relatively prime
    b) Show that there are integers r and s such that x= r^3 - 3r*(s^2) and 3*(r^2)*s-s^3=1
    c) Find all solutions in integers x^2+1=y^3 by analyzing the equations for r and s in part (b)
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    Quote Originally Posted by totally-confused View Post
    a) Show that if x and y are integers such that x^2+1=y^3, then x - i and x + i are relatively prime
    Let d be a greatest common divisor between x-i and x+i. Then d|(x-i) and d|(x+i). But since d| (x-i) it means d|[(x+i) - 2i] but d|(x+i) thus d|2i. Thus, d|i or d|2. If d|i then d is a unit, and this is what we want. Thus, it suffices to show d|2 is impossible, and the (non-unit) divisors of 2 are \pm 2,\pm i2 because they are all associate it sufficies to show d=2 is impossible. If 2|(x-i) and 2|(x+i) then 4|(x^2+1). But simple trial and error on the congruences mod 4 shows it is impossible to have a square congruent to 1 mod 4. Thus, 4\not | (x^2+1), a contradiction. Thus, d\not = 2. And so d must be a unit.
    b) Show that there are integers r and s such that x= r^3 - 3r*(s^2) and 3*(r^2)*s-s^3=1
    Here we use the fact that \mathbb{Z}[i] is a unique factorization domain. As a consequence it means if (x-i)(x+i) = y^3 and (x-i),(x+i) are relatively prime then each factor must be a cube itself, so, x+i = \alpha^3 = (r+is)^3 = (r^3 -3rs^2) + i(3r^2s - s^3) for some r,s\in \mathbb{Z}. Thus, x = r^3 - 3rs^2 and 3r^2s - s^3 = 1.
    c) Find all solutions in integers x^2+1=y^3 by analyzing the equations for r and s in part (b)[/FONT][/COLOR]
    Since 3r^2s - s^3 = 1 \implies s(3r^2 - s^2) = 1. There are two possibilities s=\pm 1 and so 3r^2 - 1 = \pm 1 so r=0 which means x=0 which leads to y=0. Thus, x=0,y=0 is the only solution.
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