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Consider an equation of the form.

$$z^3=y^2+4$$

where z ,y $\in \Bbb N$

Hence, the solution of form (z,y) are (2,2) and (5,11). So How to prove that this diophantine equation only has these two solutions.

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    $\begingroup$ These are called Mordell equations : a good link is here, along with the answer you are looking for: www.math.uconn.edu/~kconrad/blurbs/gradnumthy/mordelleqn1.pdf $\endgroup$ – астон вілла олоф мэллбэрг Oct 31 '17 at 10:12
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Suppose $z^3=y^2+4 = (2-yi)(2+yi)$, if the two factors $2-yi$ and $2+yi$ are coprime, then $(u+vi)^3 = 2+yi$, and the rest is easy. However, they need not be coprime, a more comprehensive analysis is necessary.


Let us first assume $z$ is even, letting $z=2z_1,y=2y_1$ gives $2z_1^3 = y_1^2+1$. Modulo $4$ shows that $z_1$ must be odd. Write $2z_1^3 = (y_1+i)(y_1-i)$. The two factors on the right cannot be coprime, since LHS is divisible by $2$ but not divisible by $4$. Let $a$ be their GCD, then $a|2$. Obviously $a$ cannot be $2$, so $a = 1+i$, a Gaussian prime factor of $2$. Therefore $$z_1^3 = \frac{y_1+i}{1+i} \frac{y_1-i}{1-i} = \frac{(y_1-1)+(y_1+1)i}{2}\frac{(y_1-1)-(y_1+1)i}{2}$$ Now two factors on the right are coprime, let $(u+vi)^3 = \frac{(y_1-1)+(y_1+1)i}{2}$, equating real and imaginary part and cancel $y_1$ gives $$-1 = (u+v)(u^2-4uv+v^2)$$ Hence $$\begin{cases}u+v = -1 \\ u^2-4uv+v^2 = 1 \end{cases} \quad \text{or} \quad \begin{cases}u+v = 1 \\ u^2-4uv+v^2 = -1 \end{cases}$$ Only the first case yields integral solution: $(u,v)=(-1,0)$ or $(0,-1)$. Hence plugging back gives $y_1 = \pm 1, z_1 = 1$. This gives the solution $y=\pm 2,z=2$.


If $z$ is odd. Assume $a|(2-yi), a|(2+yi)$, then $a|4$. So $a = 1,1+i,2$ or $4$. But only $a=1$ is possible, because $z^3 = (2-yi)(2+yi)$ and $z$ is odd. Hence $(2+yi) = (u+vi)^3$, which gives $(u,v) = (-1,\pm 1)$ or $(2,\pm 1)$. The first case is ruled out since $z$ is odd. The second case gives a new solution $y=\pm 11, z=5$.

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Since $z^3=y^2+4=(2-y\,i)(2+y\,i)$ and the Gaussian integers have unique prime factorization, $2+y\,i$ must be a third power, too. $(u+v\,i)^3=2+y\,i$ with integers $u,v$ means $u\,(u^2-3\,v^2)=2$, i.e. $u$ must be a divisor of $2$. There aren't very many divisors of $2$, so it's easy to see that $(u,v)=(-1,\pm1)$ and $(u,v)=(2,\pm1)$ are the only solutions.
NOTE: $2+y\,i$ and $2-y\,i$ aren't necessarily coprime, but every common prime factor must be a divisor of $(2+y\,i)+(2-y\,i)=4=-(1+i)^4$, i.e. a common factor can only be a power of $1+i$. If the exponent of $1+i$ in the factorization of $2+i\,y$ is $\alpha$, the exponent in the factorization of $(2-y\,i)(2+y\,i)$ must be $2\alpha$, since $1-i=-i\,(1+i)$, so $3|\alpha$. Any other prime factors can't be common between $2+y\,i$ and $2-y\,i$, so $2+y\,i$ must be a cube.
Of course, $z=u^2+v^2$.

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    $\begingroup$ You didn't prove $2-yi$ and $2+yi$ are coprime so you can't claim they must be a cube as well. $\endgroup$ – i9Fn Oct 31 '17 at 11:52
  • $\begingroup$ @i9Fn They don't have to be coprime for the claim to be true, see my added note. $\endgroup$ – Professor Vector Oct 31 '17 at 20:01

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