I am trying to prove that for an odd prime $p$, if $x^3+2y^3=p$ has a solution over positive integers, then it is unique.
My work,
Assume $(x,y)=(a,b)$ and $(x,y)=(c,d)$ are different solutions. $\left (\dfrac{a}{b}\right )^3\equiv \left (\dfrac{c}{d}\right )^3\equiv -2\pmod{p}$. Let $\dfrac{a}{b}\equiv r\pmod{p}$ and $\dfrac{c}{d}\equiv t\pmod{p}$. $$r^3-t^3\equiv 0\pmod{p}\Longleftrightarrow (r-t)(r^2+t^2+rt)\equiv 0\pmod{p}$$ I have proved that if $r\equiv t$, then there is a contradiction. However, I could not find a contradiction for $r^2+t^2+rt\equiv 0$. So, in other words, I have proved the question for primes such that $p\equiv 2\pmod{3}$, because $r^2+t^2+rt\equiv 0$ does not have solution for those primes (except zero). Can anyone help me for "$r^2+t^2+rt\equiv 0$" part?
I have check the question for many primes and it seems true. Yet, if you have a counterexample, that will be great.
I've found a counter example: Take $p=8317$, then $(19,9)$ and $(5,16)$ are both a valid solution with $x^3+2y^3=p$.
Another counter example is $p=1459$ with $(11,4)$ and $(1,9)$ as valid solutions.
However, $p\equiv 1 \mod 3$, so your statement could be true for those primes with $p\equiv 2 \mod 3$.