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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?

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  • $\begingroup$ Perfect question and answer (+1) $\endgroup$ – Mostafa Ayaz Nov 30 '18 at 20:36
  • $\begingroup$ 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 $f-e$... $\endgroup$ – xarles Nov 30 '18 at 20:43
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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.

Nicely done anyway.

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