how to solve the following mordell equation:$ y^2 = x^3 – 3$

i just started solving mordell’s equations and get a little bit stuck. For example:

$ y^2 = x^3 – 3$.

I know that $x$ must be odd, for if $x$ is even $y^2 \equiv 5 \pmod{8}$. So $x \equiv \{1,3\} \pmod{4}$. Now note the following if we add 4 at both sides:

$y^2 + 4 = (x+1)(x^2 -x +1)$. The right hand side is the same as $(x-1)^2 + x$

I now know $(x^2 -x +1)$ is divisible by a prime divisor $3 \pmod{4}$ if $x \equiv 3 \pmod {4}$ and divisible by $1 \pmod{4}$ if $x \equiv 1 \pmod{4}$.

So we get $y^2 + 4 \equiv 0\pmod{p}$, which is solvable if $\left(\frac{-2}{p}\right)$ = 1. (I don’t think this step is valid). this is only true if the following occurs:

$p \equiv 1 \pmod{4}$ and $p \equiv +- 1\pmod{8}$.

or

$p \equiv 3 \pmod{4}$ and $p \equiv +- 3\pmod{8}$.

so $p \equiv \{1,3\}\pmod{8}$

but $y^2 + 4$ is either $4 \pmod{8}$ or $5 \pmod {8}$, so there are no solutions?

Am i solving this the correct way or is there a shorter path?

Kees

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Yes, I think this is the best way. Keith Conrad has answered this with many examples in his article Examples of Mordell’s equation; e.g., compare with Theorem $2.2$ and $2.3$, which is very similar, for $y^2=x^3-5$ and $y^2=x^3-6$.

Your solution is flawed, so here’s a complete solution: As you said, $x$ is odd, because if $x$ were even, then $y^2\equiv -3\pmod{8}$, which is not a quadratic residue. Then $y=2k$ is even.

$4(k^2+1)=(x+1)\left(x^2-x+1\right)$.

$x^2-x+1$ is always odd. Let $p$ be a prime divisor of $x^2-x+1$. Then $p\equiv 1\pmod{4}$, because if $p\equiv 3\pmod{4}$, then $k^2\equiv -1\pmod{p}$, contradiction (by Quadratic Reciprocity). Therefore $x^2-x+1\equiv 1\pmod{4}$, so $4\mid x(x-1)$, so $4\mid x-1$, i.e. $x\equiv 1\pmod{4}$. But then $x+1\equiv 2\pmod{4}$, contradiction, because this implies $4\nmid (x+1)\left(x^2-x+1\right)$ (while $4$ divides the LHS). No solutions exist.