Integral solution of $x^2+1=y^3$

Consider $(x,y)\in\mathbb{Z}^2$ being a solution of the above equation. If $y=0$ one has $x^2+1> 0$. Let $y\in 2\mathbb{Z}\backslash \{0\}.\Rightarrow 4\mid y^3\Rightarrow x^2=y^3-1\equiv_4 3$, which contradicts $x^2\equiv_4 0,1,2$ for any $x\in\mathbb{Z}$. Therefore let $y\in 2\mathbb{Z}+1\Rightarrow x^2=y^3-1\in 2\mathbb{Z} \Rightarrow x\in 2\mathbb{Z}$, since from $x\notin 2\mathbb{Z}\Rightarrow x^2\notin 2\mathbb{Z}$.

Therefore consider solutions $(x,y)\in 2\mathbb{Z}\times (2\mathbb{Z}+1)$. In $\mathbb{Z}[i]$, which is euclidean, therefore factorial, one has $$x^2+1=(x+i)(x-i).$$

Since any unit is a divisor of any element and every element splits into a unique factorization of prime elements, consider w.l.o.g. $p\in \mathbb{Z}[i]$ prime, such that

$$p\mid (x+i)\quad\wedge\quad p\mid (x-i).$$

$\Rightarrow p \mid (x+i)-(x-i)=2i=(1+i)^2\Rightarrow p\mid (1+i)$, since $p$ is prime.

Let $\alpha\in\mathbb{Z}[i]$, such that $(1+i)=\alpha\cdot p.$

$$\Rightarrow N(\alpha)N(p)=2 \Rightarrow N(\alpha)=1,\text{ since p is prime}\Rightarrow \alpha\in\{1,-1,i,-i\}$$

It is easy to see that $(x+i),(x-i)\in2\mathbb{Z}\times (2\mathbb{Z}+1)i$ are no multiple of $\alpha(1+i)$. This means

$$p=\alpha( 1+i )\nmid x+i,\quad p=\alpha( 1+i )\nmid x-i,$$

i.e. there is no prime element $p$ which divides both $x+i$ and $x-i$. Therefore the greatest common divisor of $x+i$ and $x-i$ must be an unit.

Question: How can I derive contradiction from here? I saw someone stating that $x+i$ must be a cube in $\mathbb{Z}[i]$, but why? From having

$$(x+i)(x-i)=x^2+1=y^3$$

does not follow that $(x+i)$ is a cube right? Somehow one must use that $gcd(x+i,x-i)=1$?

• What are you trying to prove? – Hans Engler Apr 23 '18 at 19:21
• That there are no solutions except of $(x,y)=(0,1)$. – user408858 Apr 23 '18 at 19:23
• How is it easy to see that $x+i$ and $x-i$ are no multiples of $\alpha(1+i)$? – Inactive - Objecting Extremism Apr 23 '18 at 20:39
• Consider the cases 1.) $a,b\in 2\mathbb{Z}$, 2.) $a,b\in 2\mathbb{Z}+1$, 3.) $a\in 2\mathbb{Z}$, $b\in 2\mathbb{Z}+1$, 4.) $b\in 2\mathbb{Z}$, $a\in 2\mathbb{Z}+1$. Then $(a+bi)\alpha (1+i)$ is either in $2\mathbb{Z}\times 2\mathbb{Z}i$ or in $(2\mathbb{Z}+1)\times(2\mathbb{Z}+1)i$. – user408858 Apr 23 '18 at 20:43

For the full proof, see Theorem $3.2$ in Keith Conrad's notes on the Mordell curve, page $5-6$. Conrad has $x$ and $y$ interchanged, so it is $$y^2=x^3-1.$$ The only integral solutions are $(x,y)=(1,0)$.
• Thanks. I am having troubles understanding following statement: If $(y-i)$ and $(y+i)$ are relatively prime, then since their product is a cube, each factor must be a cube up to unit multiple, by unique factorization in $\mathbb{Z}[i]$. – user408858 Apr 23 '18 at 19:28
• This is just unique factorization, like in $\mathbb{Z}$. For a completely elementary proof see here. – Dietrich Burde Apr 23 '18 at 19:30
• you have $ab=x^3$ with $\gcd(a,b)=1$ so if $p$ is a prime divisor of $x$ then it divides either $a$ or $b$. Let suppose it is $a$. But then $p^3\tilde x^3=p\tilde ab$ and again $p$ divides $\tilde a$, and so on. In the end $p^3$ divides $a$. – zwim Apr 23 '18 at 19:37
• @zwim: Isn't this only true, if $x$ is prime? Why should $x^3$ be already the unique prime factorization of itself? I'm sorry. I think I am overseeing something really really simple... – user408858 Apr 23 '18 at 19:37
• @zwim Ok, I see. Then since there are no zero divisors in $\mathbb{Z}[i]$, it holds $p^2\tilde{x}^3=\tilde{a}b$. Since $p$ was no divisor of $b$ because of $gcd(a,b)=1$, $p\mid \tilde{a}$ and $p^2\tilde{\tilde{a}}=p\tilde{a}=a$. And again the last step. Now I have it. Thank you so much! – user408858 Apr 23 '18 at 19:52