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Thread: 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 $\displaystyle d$ be a greatest common divisor between $\displaystyle x-i$ and $\displaystyle x+i$. Then $\displaystyle d|(x-i)$ and $\displaystyle d|(x+i)$. But since $\displaystyle d| (x-i)$ it means $\displaystyle d|[(x+i) - 2i]$ but $\displaystyle d|(x+i)$ thus $\displaystyle d|2i$. Thus, $\displaystyle d|i$ or $\displaystyle d|2$. If $\displaystyle d|i$ then $\displaystyle d$ is a unit, and this is what we want. Thus, it suffices to show $\displaystyle d|2$ is impossible, and the (non-unit) divisors of $\displaystyle 2$ are $\displaystyle \pm 2,\pm i2$ because they are all associate it sufficies to show $\displaystyle d=2$ is impossible. If $\displaystyle 2|(x-i)$ and $\displaystyle 2|(x+i)$ then $\displaystyle 4|(x^2+1)$. But simple trial and error on the congruences mod 4 shows it is impossible to have a square congruent to 1 mod 4. Thus, $\displaystyle 4\not | (x^2+1)$, a contradiction. Thus, $\displaystyle d\not = 2$. And so $\displaystyle d$ 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 $\displaystyle \mathbb{Z}[i]$ is a unique factorization domain. As a consequence it means if $\displaystyle (x-i)(x+i) = y^3$ and $\displaystyle (x-i),(x+i)$ are relatively prime then each factor must be a cube itself, so, $\displaystyle x+i = \alpha^3 = (r+is)^3 = (r^3 -3rs^2) + i(3r^2s - s^3)$ for some $\displaystyle r,s\in \mathbb{Z}$. Thus, $\displaystyle x = r^3 - 3rs^2$ and $\displaystyle 3r^2s - s^3 = 1$.
    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 $\displaystyle 3r^2s - s^3 = 1 \implies s(3r^2 - s^2) = 1$. There are two possibilities $\displaystyle s=\pm 1$ and so $\displaystyle 3r^2 - 1 = \pm 1$ so $\displaystyle r=0$ which means $\displaystyle x=0$ which leads to $\displaystyle y=0$. Thus, $\displaystyle x=0,y=0$ is the only solution.
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