Solving Diophantine Equation in Z[i] - x^2 + 4 = y^3

[Galileo]

This problem is working on my nerves. I`m trying to find all integer solutions to the equation $x^2+4=y^3$ using the PID of Gaussian integers Z.

My thoughts.
By inspection (2,2) is a solution.
Suppose (x,y) is a solution. I write the equation as $(x+2i)(x-2i)=y^3$.
I now look at the ideal (x+2i,x-2i)=(d) with d a generator. d divides x+2i and x-2i, so it also divides the difference 4i.

What I want is to find conditions under which x+2i and x-2i are coprime in Z. Then I can show that (under the conditions) x+2i has to be a third power in Z and that no solutions exist (under this condition).

Any help is appreciated.

 

[Galileo]

Last cry for help...

 

[shmoe]

This may be too late, but anyways:

If d is a divisor of of x+2i and x-2i, then d divides 4i like you said. What can you then say about the norm of d? If x is odd, what does this say about the norm of x+2i?

If x is even, consider the original equation mod 8. You should be able to reduce it to something easier to handle.

 

[Galileo]

Thanks shmoe.
Here's what I came up with.
Let (x,y) be a solution. Because d|4i we have N(d)|N(4i)=16=2^4. So N(d) is 1,2,4,8 or 16. We also have d|x+2i so N(d)|x^2+4.
Suppose x is odd, then x^2+4 is odd and N(d) must be 1 so d is a unit. Then (d)=Z and (x+2i) and (x-2i) are coprime. Then x+2i and x-2i will not have any common irreducible factors, so they must both be equal to a third power in Z, because every unit is too and their product is y^3.
We then get the equation:
$$x+2i=(a+bi)^3=a(a^2-3b^2)+(3a^2-b^2)bi$$
If $b=1$, then $3a^2-1=2$ so $a=\pm 1$, yielding $x=\pm 2$, giving the solutions (2,2) and (-2,2).
(Although I should not count these, because I assumed x is odd )
For b=-1 and b=2 there are no solutions, but for b=-2 I get $x=\pm 11$ with the solutions (11,5) and (-11,5)

If x is even then y must be even and thus y^3 congruent 0 mod 8. So $x^2 \equiv 4 \pmod 8$. So x=2 or x=6 (mod 8), but I`m not sure what to do with that.

 

[shmoe]

If $b=1$, then $3a^2-1=2$ so $a=\pm 1$, yielding $x=\pm 2$, giving the solutions (2,2) and (-2,2).
(Although I should not count these, because I assumed x is odd )
For b=-1 and b=2 there are no solutions, but for b=-2 I get $x=\pm 11$ with the solutions (11,5) and (-11,5)

Correct. The even solutions will appear again, don't worry.

If x is even then y must be even and thus y^3 congruent 0 mod 8. So $x^2 \equiv 4 \pmod 8$. So x=2 or x=6 (mod 8), but I`m not sure what to do with that.

You can write x=2a, where a is odd, and y=2b. Stuff into your equation and what can you say?

 

[Galileo]

Correct. The even solutions will appear again, don't worry.
You can write x=2a, where a is odd, and y=2b. Stuff into your equation and what can you say?

Stuffing... I get $a^2+1=2b^3$.
So I'd say I get another diophantine equation
Well, at least I know a is odd, from which follows that b must also be odd. If b was even the right side would be 0 mod 4 while the left is 2 mod 4.

After puzzling I feel like I`m reducing possible solutions, but I still have an infinite number of options. I may be going the wrong way, but I let a=2m+1 and b=2k+1 and got: m(m+1)=k(4k^2+6k+3). Two solutions are ofcourse k=m=0 and k=0, m=-1, corresponding to (x,y)=(2,2) and (x,y)=(-2,2).

 

[shmoe]

From this point: $a^2+1=2b^3$ you can do some factoring over Z. Try to get something like u^2+v^2=b^3. It might help to notice this is expressing b^3 as the norm of an element in Z.