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I want to solve the Diophantine equation $y^3 = 4x^2+4x+ 5$ for $x,y \in \mathbb{Z}$.

The right hand side factors as $(2x+1-2i)(2x+1+2i)$. Am I right that such a factorization can be found using the quadratic formula?

Now I think that the equation can be solved by performing a descent as $\mathbb{Z}[i]$ is a UFD. But for doing that we need that the above two factors are coprime. Are the above two factors coprime, and how do I show that?

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  • $\begingroup$ The two factors are degree 1 polynomials, so they can only be factored into a product of a degree 0 and degree 1 polynomial, i.e. the best you can do is factor out a constant. The question reduces to finding $\text{gcd}(2,1-2i)$ and $\text{gcd}(2,1+2i)$ over $\mathbb{Z}[i]$ and showing they are coprime. $\endgroup$ – AlexanderJ93 Dec 7 '18 at 21:27
  • $\begingroup$ I can see that $(x,y)=(5,5)$ and $(x,y)=(-6,5)$ are solutions. I don't know if there are more, but don't think that there are more. $\endgroup$ – Batominovski Dec 7 '18 at 21:48
  • $\begingroup$ Another method would probably be viewing this as an elliptic curve: the substitution $x\to{}(x-1)/2$ for odd $x$, will give $y^3-5=4(x^2-1)/4$ and thus $y^3-4=x^2$. Now this is an elliptic curve with a finite number of integer solutions that can be found by the Nagell Lutz theorem. $\endgroup$ – Μάρκος Καραμέρης Dec 7 '18 at 21:58
  • $\begingroup$ @AlexanderJ93 I don't understand how you get to the $\gcd$'s? And where did $x$ go? $\endgroup$ – Jens Wagemaker Dec 7 '18 at 22:02
  • $\begingroup$ @JensWagemaker You cannot factor linear equations any more except by factoring out constants. The only way you can factor out constants is if the $\text{gcd}$ of all coefficients is more than 1. If $\gcd(2,1-2i) = a>1$, then you can factor $a$ out of $(2x+1-2i)$, similarly for $\gcd(2,1+2i) = b>1$. Then if you can factor $a$ and $b$ out of the two terms, respectively, and $a$ and $b$ are not coprime, then the original factors were not coprime. Otherwise, if $a$ and $b$ are coprime and $(2x+1-2i)/a \neq (2x+1+2i)/b$, then all factors are distinct and so they are coprime. $\endgroup$ – AlexanderJ93 Dec 7 '18 at 22:11
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Let $x,y\in\mathbb{Z}$ be such that $y^3=4x^2+4x+5=(2x+1-2\text{i})(2x+1+2\text{i})$. Let $d\in\mathbb{Z}[\text{i}]$ be a greatest common divisor of $2x+1-2\text{i}$ and $2x+1+2\text{i}$. Thus, $d$ divides the difference $4\text{i}$, but the norm of $d$ must be odd (as the norm of $d$ divides the odd integer $4x^2+4x+5$). This shows that the norm of $d$ is an odd number dividing the norm of $4\text{i}$, which is $16$. Hence, $d$ must have a unit norm, whence it is a unit. Ergo, $\gcd(2x+1-2\text{i},2x+1+2\text{i})=1$ in $\mathbb{Z}[\text{i}]$.

Since $\gcd(2x+1-2\text{i},2x+1+2\text{i})=1$ and the units of $\mathbb{Z}[\text{i}]$ are perfect cubes in $\mathbb{Z}[\text{i}]$, both $2x+1-2\text{i}$ and $2x+1+2\text{i}$ must be perfect cubes in $\mathbb{Z}[\text{i}]$. That is, for some $u,v\in\mathbb{Z}$, $$2x+1+2\text{i}=(u+v\text{i})^3\,,$$ which implies $$2=(3u^2-v^2)v\,.$$ Consequently, $v\in\{\pm1,\pm2\}$. It is easy to see that only $(u,v)=(\pm1,1)$ and $(u,v)=(\pm1,-2)$ works. This gives $$2x+1=u(u^2-3v^2)\in\{\pm2,\pm11\}\,.$$ That is, $x=5$ or $x=-6$, leading to the two solutions $(x,y)=(5,5)$ and $(x,y)=(-6,5)$.

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