Problem 1351

One can verify that $x^3 \equiv 0, 1, -1 \pmod{7}$. Since $2^{2009} = (2^3)^{669} \cdot 2^2 \equiv 1^{669} \cdot 4 \equiv 4 \pmod{7}$ and $x^3 + y^3 \equiv -2, -1, 0, 1, 2 \pmod{7}$, it is not possible that $x^3 + y^3 \equiv 2^{2009} \pmod{7}$, 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 $x = d \cdot x'$ and $y = d \cdot y'$, where $d = \gcd(x, y)$. So $x^3 + y^3 = d^3(x'^3 + y'^3) = d^3((x'+y')(x'^2 - x'y'+y'^2))$ and one can prove that $\gcd(x'+y', x'^2-x'y'+y'^2) = \gcd(x'+y', (x'+y')^2 - 3x'y') = \gcd(x'+y', 3x'y')$ equals 1 or 3. Since $x^3+y^3$ is not divisible by 3, $\gcd(x' + y', x'^2 - x'y' + y'^2) = 1$ and, considering that both $x' + y'$ and $x'^2 - x'y' + y'^2$ are powers of 2, $x' + y' = 1$, which is not possible because $x'$, $y'$ are positive integers.