Problem 1152

Solution. Let us assume that $n^{5}+n^{3}+2 n^{2}+2 n+2$ and $2 n^{2}+n+2$ are perfect cubes, i.e., there exist natural numbers $x$ and $y$ such that

$$\begin{aligned} & n^{5}+n^{3}+2 n^{2}+2 n+2=x^{3} \\ & 2 n^{2}+n+2=y^{3} \end{aligned}$$

Then

$$x^{3}-y^{3}=n^{5}+n^{3}+n=n\left(n^{4}+2 n^{2}+1-n^{2}\right)=n\left(n^{2}+n+1\right)\left(n^{2}-n+1\right)$$

so

$$(x-y)\left(x^{2}+x y+y^{2}\right)=n\left(n^{2}+n+1\right)\left(n^{2}-n+1\right)$$

The product on the right side of the last equation is divisible by 3. Indeed, if $n=3 k$ then the first factor is divisible by 3, if $n=3 k+1$, then the second factor is divisible by 3, and for $n=3 k+2$ the third factor is divisible by 3. Therefore, $3 \mid x-y$ or $3 \mid x^{2}+x y+y^{2}$. However, from $x^{2}+x y+y^{2}=(x-y)^{2}+3 x y$ it follows that $3 \mid x^{2}+x y+y^{2}$ if and only if $3 \mid x-y$. Therefore, in this case, $3 \mid x-y$ and also $3 \mid x^{2}+x y+y^{2}$, which means that $9 \mid x^{3}-y^{3}$.

From the above considerations, it follows that $9 \mid n\left(n^{2}+n+1\right)\left(n^{2}-n+1\right)$. Furthermore, no two of the numbers $n, n^{2}+n+1$, and $n^{2}-n+1$ are simultaneously divisible by 3 (Why?), so the following cases are possible.

Case 1. If $9 \mid n$, then $x^{3}=n^{5}+n^{3}+2 n^{2}+2 n+2 \equiv 2(\bmod 9)$, which is not possible since the cube of a natural number when divided by 9 does not give a remainder of 2.

Case 2. If $9 \mid n^{2}+n+1$, then $a(a+1)=a^{2}+a \equiv 8(\bmod 9)$, but a direct check shows that the product of two consecutive natural numbers when divided by 9 does not give a remainder of 8.

Case 3. If $9 \mid n^{2}-n+1$, then $a(a-1)=a^{2}-a \equiv 8(\bmod 9)$, but a direct check shows that the product of two consecutive natural numbers when divided by 9 does not give a remainder of 8.