Belorusija 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 $$(x^2+5)(2x+1)=2\cdot p^n.(1)$$ Since $(2x+1)$ is odd for all natural $x$, we have $p\ne 2$ 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, $(x^2+5)÷(2x+1)$. 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 $p$.