, 2012

Answer: $(x; n; p) = (1; 2; 3)$, $(x; n; p) = (3, 2, 7)$.
It is easy to see that $2x^3 + x^2 + 10x + 5 = (x^2 + 5)(2x + 1)$, so the initial equality can be rewritten as
$$(x^2 + 5)(2x + 1) = 2 \cdot p^n. \quad (1)$$
Since $(2x + 1)$ is odd for all natural $x$, we have $p \neq 2$ and $2x + 1 = p^k$, $x^2 + 5 = 2 \cdot p^{n-k}$. It is evident that $x^2 + 5 > 2x + 1$ for all $x \in \mathbb{N}$ and $p \ge 3$, then $n - k \ge k$. Therefore, $(x^2 + 5) \div (2x + 1)$. So the number $(x^2 + 5)/(2x + 1)$ is integer, but then the number $2(x^2 + 5)/(2x + 1)$ is also integer. Since $2(x^2 + 5) = x(2x + 1) + (10 - x)$, we see that $(10 - x)/(2x + 1)$ must be also integer, and so the number $2(10 - x)/(2x + 1)$ must be integer too. But $2(10 - x) = -(2x + 1) + 21$, so the number $21/(2x + 1)$ must be integer. It is possible only if $2x+1 = 1, 3, 7, 21$, i.e. if $x = 1, x = 3, x = 10$.

For $x = 1$ we have $(x^2 + 5)(2x + 1) = 6 \cdot 3 = 18 = 2 \cdot 3^2$, so $p = 3$, $n = 2$.
For $x = 3$ we have $(x^2 + 5)(2x + 1) = 14 \cdot 7 = 2 \cdot 7^2$, so $p = 7$, $n = 2$.
For $x = 10$ we have $(x^2 + 5)(2x + 1) = 405 \cdot 21 = 5 \cdot 7^2 \cdot 3^5$, i.e. this number cannot be $2 \cdot p^n$ for any natural $n$ and any prime $p$.