# Solving the Diophantine equation $y^2 = 4x^3 + 1$ for $x,y \in \mathbb{Z}$

I want to solve the Diophantine equation $$y^2 = 4x^3 + 1$$ for $$x,y \in \mathbb{Z}$$.

Note that $$y$$ is odd, since $$y$$ even would give a contradiction $$\mod{2}$$. Hence $$\frac{y-1}{2}, \frac{y+1}{2} \in \mathbb{Z}$$.

So we can rewrite the equation to: $$\frac{y-1}{2} \frac{y+1}{2} = x^3$$.

Claim $$\frac{y-1}{2}$$ and $$\frac{y+1}{2}$$ are coprime.

Proof: Let $$d = \gcd(\frac{y-1}{2},\frac{y+1}{2})$$. Let $$\frac{y-1}{2} = a d$$ and $$\frac{y+1}{2} = b d$$ for some $$a,b \in \mathbb{Z}$$. Then $$y + 1 = bd \,2$$ and $$y-1 = a d\, 2$$, so $$y + 1 = ad\,2+2$$, it follows that $$bd = ad + 1$$, so $$d(b-a)=1$$, so $$d =\pm1$$. So we conclude that indeed $$\frac{y-1}{2}$$ and $$\frac{y+1}{2}$$ are coprime.

Since they are coprime we can perform a descent. That gives $$\frac{y-1}{2} = e^3$$ and $$\frac{y+1}{2} = f^3$$ for some coprime $$e,f \in \mathbb{Z}$$. After adding and subtracting these two equations we find that $$y = e^3+f^3$$ and $$1 = f^3 - e^3$$.

Am I now correct to say that the only solutions are $$f = 1, e=0$$ and $$f = 0, e=-1$$?

This gives $$y=\pm1$$. In the above equation we see that in both cases $$x = 0$$.

We conclude by saying that the only solutions are $$(x,y) = (0,\pm1)$$.

Did I make any mistakes?

Are there more efficient methods?

• Perfect question and answer (+1) – Mostafa Ayaz Nov 30 '18 at 20:36
• May be you could explain why $1=f^3-e^3$ has only the two solutions you give. For example using that $f^3-e^3$ is divisible by $f-e$... – xarles Nov 30 '18 at 20:43

Of course you can, from $$f^3-e^3=1$$ we get $$(f-e)(f^2+fe+e^2)=1$$
so $$f-e=1$$ and $$f^2+fe+e^2 = 1$$ or $$f-e=-1$$ and $$f^2+fe+e^2 = -1$$
So you have 2 cases. You can express $$f$$ with $$e$$ and plug in the second equation...
Else, I would introduce $$y= 2k+1$$ for some $$k$$, then you would have $$k(k+1)=x^3$$ which is a little easer to handle. One thing you can notice faster is that $$k$$ and $$k+1$$ are coprime.