XXXI Brazilian Math Olympiad

One can verify that $x^3\equiv 0,1,-1(\mathrm{mod}7)$. Since $2^{2009}=(2^3{)}^{669}\cdot 2^2\equiv 1^{669}\cdot 4\equiv 4(\mathrm{mod}7)$ and $x^3+y^3\equiv -2,-1,0,1,2(\mathrm{mod}7)$, it is not possible that $x^3+y^3\equiv 2^{2009}(\mathrm{mod}7)$, so the equation $x^3+y^3=2^{2009}$ has no solutions.

Comment: One can solve the problem without the aid of arithmetic mod $n$: let $x=d\cdot x^{\prime }$ and $y=d\cdot y^{\prime }$, where $d=\gcd (x,y)$. So $x^3+y^3=d^3(x^{\prime 3}+y^{\prime 3})=d^3((x^{\prime }+y^{\prime })(x^{\prime 2}-x^{\prime }{y}^{\prime }+y^{\prime 2}))$ and one can prove that $\gcd (x^{\prime }+y^{\prime },x^{\prime 2}-x^{\prime }{y}^{\prime }+y^{\prime 2})=\gcd (x^{\prime }+y^{\prime },(x^{\prime }+y^{\prime }{)}^2-3x^{\prime }{y}^{\prime })=\gcd (x^{\prime }+y^{\prime },3x^{\prime }{y}^{\prime })$ equals 1 or 3. Since $x^3+y^3$ is not divisible by 3, $\gcd (x^{\prime }+y^{\prime },x^{\prime 2}-x^{\prime }{y}^{\prime }+y^{\prime 2})=1$ and, considering that both $x^{\prime }+y^{\prime }$ and $x^{\prime 2}-x^{\prime }{y}^{\prime }+y^{\prime 2}$ are powers of 2, $x^{\prime }+y^{\prime }=1$, which is not possible because $x^{\prime }$, $y^{\prime }$ are positive integers.