# Solving the Diophantine equation $y^3 = 4x^2+4x+ 5$ for $x,y \in \mathbb{Z}$

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?

• 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. – AlexanderJ93 Dec 7 '18 at 21:27
• 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. – Batominovski Dec 7 '18 at 21:48
• 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. – Μάρκος Καραμέρης Dec 7 '18 at 21:58
• @AlexanderJ93 I don't understand how you get to the $\gcd$'s? And where did $x$ go? – Jens Wagemaker Dec 7 '18 at 22:02
• @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. – AlexanderJ93 Dec 7 '18 at 22:11

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 $$(x,y)=(-6,5)$$.