Does the equation $x^3-y^3=b$ has solutions in a finite field $F$ where $b\in F$?
I know it is true for the equation $$x^2-y^2=b\,.$$ As for the case $\text{char}\ F \neq 2$, $$x^2-y^2=(x+y)(x-y)=b\cdot 1\,,$$ so $x+y=b$ and $x-y=1$ has simultaneous solutions in finite field $F$ for any $b \in F$ namely $$x=\frac{b+1}{2}\text{ and } y=\frac{b-1}{2}\,.$$ The case with $\text{char}\ F=$ $2$ is easy as every element in $F$ is square so put $y=0$ and we can find $x$ easily.
But what about $x^3-y^3=(x-y)(x^2+y^2+xy)=b\cdot 1$?
Here it gets complicated a bit, as we do not have two linear equations to solve here. Which subject deals with such questions, I guess algebraic geometry or elliptic curves, I have never had any introduction to any of them, so if someone can suggest a way to attack such question without going too much in those subjects and using field arithmetic mostly, then it will be a lot helpful. Thanks!
The following theorem can be found in the paper Multiplicative Subgroups of Index $3$ in a field. This paper can be accessed here. The purpose of the paper is to prove the following theorem.
Note that $G+G$ is the same thing as $G-G$ in our case since $-1$ is a cube.
The other theorem is
That paper also talks a bit about this and other classical methods can also be used to tackle this problem such as Jacobi sums. Now we look at fields of characteristic $0$. Using Van der Waerden's theorem, Berrizbeitia proved:
The really heavy tools are needed by the authors of this paper. Using Ramsey Theory and the fact that there exists a finitely additive invariant probability measure on an ammenable group (like $K^{*}$), the authors were able to prove the analogue of the above state for infinite fields with finite characteristic.