Statement: If $x,y,n$ are odd integers such that $\gcd(x,y)=1$ and $x^2+y^2-xy=n^3$, then $x+y=2c^3-2cd^2$, where $c,d$ are co-prime integers.
Two examples for clarity:
$x=-71, y=181$
$(-71)^2+181^2-(-71)(181)=37^3$
And $x+y=181-71=2(-5)^3-2(-5)(6)^2=2c^3-2cd^2$
$x=1, y=19 \implies 1+19^2-19=7^3$ and $20=2(-2)^3-2(-2)(3)^2$
I know one way to prove this (I think) by writing $x^2+y^2-xy$ in the form $a^2+3b^2$ and proceeding from there. I’m wondering if there could be another way to prove it over integers?
I’m not 100% sure the statement is true, and I wouldn’t know where to start on a different proof. I’m curious if anyone knows of another proof, or some hints where to start. Or if the statement is false.
Also related you can have $x^2+y^2-xy=3n^3$ then it’s the case that $x+y=18e^3-18ef^2$. I wonder if this would follow from the first statement?