If $x, y$ is a solution then so is $y-x$, $-x$. Hence also $-y$, $x-y$. If the first two are the same, then $y = -x$, and $x = y-x = -2x$, so $x = y = 0$, which is impossible, since $n >0$. Similarly, if any other pair are the same. $2891 = 2 (mod.9)$ and there is no solution to $x^3 - 3xy^2 + y^3 = 2 (mod 9)$. The two cubes are each $-1$, $0$ or $1$, and the other term is $0$, $3$ or $6$, so the only solution is to have the cubes congruent to $1$ and the other term congruent to $0$. But the other term cannot be congruent to $0$, unless one of $x$, $y$ is a multiple of $3$, in which case its cube is congruent to $0$, not $1$ or $-1$.

$ (x + \omega y)^3 = x^3 - 3xy^2 + y^3 + \omega(3x^2y - 3xy^2)$ But $ a^3 = (a\omega)^3 = (a\omega^2)^3$, so, if there is a solution, we obtain other two. For $ n = 2891$, we see the equation $ mod 11$. We must have $ x^3 - 3xy^2 + y^3 \equiv 2891 \equiv 0$. If $ 11 \nmid x, y$ there is a solution for the equation $ t^3 - 3t + 1 \equiv 0$ (take $ t = x.y^{-1}$). Testing cases, we see this is impossible. So $ 11 \mid x, y$. This implies $ 11^3|x^3 - 3xy^2 + y^3 = 2891$, contradiction.

Feliz's solution is false, since 11 doesn't divide 2891...

What's the motivation behind the transformation $(x,y)\to (y-x,-x)$? I understand why one would want to go from $(x,y)\to (ax+by,cx+dy)$, but is there a better way to find $a,b,c,d$ instead of plugging that in, equating coefficients, and solving a system of equations?

claserken wrote: What's the motivation behind the transformation $(x,y)\to (y-x,-x)$? I understand why one would want to go from $(x,y)\to (ax+by,cx+dy)$, but is there a better way to find $a,b,c,d$ instead of plugging that in, equating coefficients, and solving a system of equations? Just write $y=x+k$ and see what happens.

I would like to show another way to prove the second part of the problem and to do this I will prove more general result In fact there is no solution to $n=4k+2$ suppose to the contrary that there is a solution $(x,y)$ then discuss the following cases: 1) $x$ is even then clearly $y$ is even and Hence : $(x,y)=(2a,2b)$ so $8(a^3-3ab^2+b^3)=4k+2$ which is wrong. 2) $x$ is odd and Hence: a) $y$ is even then $n$ must be odd which is wrong. b) $y$ is odd wthen $n$ must be odd which is wrong. so when $n=4k+2$ no solution but $1982=4*495+2$ Hence no sloution and we are done

Detailed Solution as #2: WLOG, Assume that $x=y+m$, Then we have, $$(y+m)^3-3y^2(y+m)+y^3=n$$$$m^3-3y^2m+(-y)^3=n$$So, we see that if $(x,y)$ is solution then $(-y,x-y)$ is also a solution. Similarly, letting $y=x+k$ gives. $$k^3-3x^2k+(-x)^3=n$$So, $(-x,y-x)$ is also a solution. Claim 1: All the solution pairs are distinct assume to the contrary that $(x,y)=(-y,x-y)\implies y=-2y \implies x=y=0$. Contradiction since $n$ is a positive integer. This proves our claim.So, we must have three different solution. [Side note: In the case of cubes check $\pmod{7} $ or $\pmod{9}$ for relevant results] Suppose for the sake of contradiction there exists a solution pair $(x,y)$ in integers for equation: $$x^3-3xy^2+y^3=2891$$We have, $x^3 \equiv 0,1,8 \pmod{9}$, Now we work on the basis of: $2891 \equiv 2 \pmod{9}$. Case I: $x,y$ are multiple of $3$. $$x^3-3xy^2+y^3 \equiv 0 \pmod{9}$$Contradiction. Case II: $x$ or $y$ is multiple of three. $$\implies x^3-3xy^2+y^3 \equiv 0 \pmod{9}$$Contradiction. Case III: $x,y$ are integers not multiple of 3. $$\implies x^3-3xy^2+y^3 \equiv 3,6 \pmod {9}$$Contradiction. So, there are no solutions when $n=2891$. And, we are done.$\blacksquare$

We can use quadratic residues for solving this.

For the first part of the problem, notice that if $(x, y)$ is a solution, then $(y-x, -x)$ and $(-y, x-y)$ are also solutions, which means that all solutions come in distinct triplets of three, which is enough. For the second part, notice that $2891 = 7^2 \cdot 59$. Now, setting $a=yx^{-1} \pmod 7$, we have $$a^3 - 3a^2 + 1 \equiv 0 \pmod 7.$$If $a^3 \equiv 1 \pmod 7$, then $a^2 \equiv 3 \pmod 7$ is not a quadratic residue. If $a^3 \equiv -1 \pmod 7$, then $3a^2 \equiv 0 \pmod 7$, contradiction as $7 \nmid a$. This shows that there are no solutions.

So do you really want the motivation? It can be factorized into $(x-y\cdot2\cos40^\circ)(x-y\cdot2\cos80^\circ)(x+y\cdot2\cos20^\circ)$. If $x-y\cdot2\cos40^\circ=0$, then $x:y=\sin80^\circ:\sin40^\circ$, and $x:y:(x-y)=\sin80^\circ:\sin40^\circ:\sin20^\circ$. Therefore, if $f(x,y)=0$, then $f(x-y,x)=f(y,y-x)=0$, and since the minimal polynomials of $x-y\cdot2\cos40^\circ$, etc. are cubic, they must differ by at most a constant factor.