62nd Belarusian Mathematical Olympiad

Answer: $(x;n;p)=(1;2;3)$, $(x;n;p)=(3;2;5)$.

It is easy to see that $x^3+3x+14=(x+2)(x^2-2x+7)$, so the initial equality can be rewritten as $$(x+2)(x^2-2x+7)=2\cdot p^n(1)$$ It is evident that $x^2-2x+7>x+2$ for all $x\in \mathbb{N}$. So for the case $x+2=2\cdot p^k$, $x^2-2x+7=2\cdot p^{n-k}$ we have $n-k\ge k\ge 0$. It means that for both the cases $2(x^2-2x+7):(x+2)$. Therefore, the number $2(x^2-2x+7)/(x+2)$ is integer, but since $2(x^2-2x+7)=2x(x+2)-8(x+2)-30$, the number $30/(x+2)$ must be integer. It possible only if $(x+2)$ is a divisor of $30$. From (1) it follows that the number $x+2$ has at most two prime divisors, one of them being $2$. So for the natural number $x$ the number $x+2$ can admit only the following values $3,5,6,10$, i.e. $x$ can admit only the values $1,3,4,8$.

For $x=1$ we have $(x+2)(x^2-2x+7)=3\cdot 6=18=2\cdot 3^2$, so $p=3$, $n=2$.

For $x=3$ we have $(x+2)(x^2-2x+7)=5\cdot 10=2\cdot 5^2$, so $p=5$, $n=2$.

For $x=4$ we have $(x+2)(x^2-2x+7)=6\cdot 15=2\cdot 3^2\cdot 5$, i.e. this number cannot be $2\cdot p^n$ for any natural $n$ and any prime $p$.

For $x=8$ we have $(x+2)(x^2-2x+7)=10\cdot 55=2\cdot 5^2\cdot 11$, i.e. this number cannot be $2\cdot p^n$ for any natural $n$ and any prime $p$.