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Here's a question from an old examination paper:

  1. Find all $(x,y)$ in $\mathbf{Z}^{2}$ where $y$ is odd and $y^2=x^3-4$.

  2. Find all $(x,y)$ in $\mathbf{Z}^{2}$ with $y$ even and $y^2=x^3 -4$.

  3. When $(x,y)$ in $\mathbf{Z}^{2}$ where $y=2Y$ is even and $y^2=x^3-4$, show that $x=2X$ with $X, Y$ odd and that $\gcd(Y+i,Y-i) = 1+i$.

An older student who has taken the exam already told us that we should look at $\mathbf{Z}[i]$ but I don't see where to go with this information. Help is greatly appreciated.

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up vote 5 down vote accepted

Write the equation as $x^3=y^2+4$. In $\mathbb Z[i]$, you have $y^2+4 = (y+2i)(y-2i)$. Any common divisor of $(y+2i)$ and $(y-2i)$ must be a divisor of $4i$. From that and the unique factorization in $\mathbb Z[i]$ you can conclude that most divisors of $y\pm 2i$ are cubes. Now ask yourself when $(a+b i)^3$ has imaginary part equal to $2$. This should get you started. I haven't checked the details, though.

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The details work. They are given on pages 397-8 of Uspensky and Heaslet, Elementary Number Theory. – Gerry Myerson Nov 9 '11 at 4:53
BTW, this equation is a Mordell equation. – lhf Nov 9 '11 at 10:54
Thanks lhf and Srivatsan for editing,too. $(a+bi)^{3} = a^{3}-3ab^{2} + i (3a^{2}b-b^{3})0$, so $Im((a+bi)^{3}) = 2 \Leftrightarrow b(3a^{2}-b^{2}=2$ and this is true for $a=\pm 1 , b =-2$ and $a=\pm 1 , b = 1 $. Apart from not knowing how to show that these solutions are unique, I don't see what to conclude from this either. Further help is greatly appreciated. – PumaDAce Nov 9 '11 at 12:05
@PumaDAce, my point is that if the product of two numbers is a cube then the two numbers must be cubes themselves. But this holds only if they are coprime, hence my comment on the common divisors. – lhf Nov 9 '11 at 12:29
I can solve 1 and 2 with your hints lhf. Do you have a clue how to go at 3), too? Thank you a lot. – PumaDAce Nov 9 '11 at 12:47

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