9.5. Find all integer solutions of the equation: $x^{3}+y^{3}=2^{30}$. Answer. $\left(0,2^{10}\right),\left(2^{10}, 0\right)$.
Solution. We will show that if $x^{3}+y^{3}$ is a power of two not less than the second, then both variables must be even numbers. Indeed, $x^{3}+y^{3}=(x+y)\left(x^{2}-x y+y^{2}\right)=2^{n}, n \geq 2$, if one of the variables is odd, then the other is also odd, since the sum of their cubes is even. In this case, the second bracket in the factorization will be an odd integer, dividing a power of two, that is, $x^{2}-x y+y^{2}= \pm 1$. Consider the last equality as a quadratic equation in terms of $x$, if it has a solution, its discriminant, equal to $-3 y^{2} \pm 4$ must be non-negative, which is possible only when $y=0, \pm 1$. For $y=0$ we get $x^{2}= \pm 1$, from which $x= \pm 1$. For $y=1$ we get $x^{2}-x+1= \pm 1$, from which $x=1,0$. For $y=-1$ we get $x^{2}+x+1= \pm 1$, from which $x=-1,0$. For odd $x, y$ we get two pairs of values: $(1,1)$ and ( $-1,-1$ ), for which $x^{3}+y^{3}$ is, respectively, 2 and -2, so only the first pair $(1,1)$ fits, and $x^{3}+y^{3}$ in this case is 2 - a power of two below the second.

Now consider the original equation. Given the proven fact, both variables are even, dividing the entire equation by 8, we get a new equation in integers $\left(\frac{x}{2}\right)^{3}+\left(\frac{y}{2}\right)^{3}=2^{27}$, in which the right-hand side is again a power of two higher than the first. Continuing in the same vein, we will reach the equation $\left(\frac{x}{2^{10}}\right)^{3}+\left(\frac{y}{2^{10}}\right)^{3}=1$. From the solution of the equation $x^{2}-x y+y^{2}= \pm 1$ in the previous part, taking into account $\left(\frac{x}{2^{10}}\right)^{3}+\left(\frac{y}{2^{10}}\right)^{3}=1$ for $\left(\frac{x}{2^{10}}, \frac{y}{2^{10}}\right)$ we get $(1,0)$ or $(0,1)$, from which $(x, y)$ is $(x, y)$ or $\left(0,2^{10}\right)$.

Grading criteria. Only guessed solutions: 0 points. Shown that if $x^{3}+y^{3}$ is a power of two not less than the second, then both variables must be even numbers: 3 points. The idea of ten divisions of the values of the variables by 2 is used: 3 points. Solution of the equation $\left(\frac{x}{2^{10}}\right)^{3}+\left(\frac{y}{2^{10}}\right)^{3}=1: 1$ point.