<s>[INST] How find this $x^3-5x+10=2^y$ let $x,y$ is positive integer,and such
$$x^3-5x+10=2^y$$
find all $x,y$.
since $$x=1\Longrightarrow 1^3-5+10=6$$ can't
$$x=2,2^3-5\cdot 2+10=8=2^3$$
so $x=2,y=3$ 
$$x=3,LHS=27-15+10=22$$
$$x=4,LHS=64-20+10=54$$
$$x=5,LHS=125-25+10=110$$
$$x=6,LHS=216-30+10=236$$
$$\cdots$$
I find $$(x,y)=(2,3)$$
I only find $x\le 7$ this solution.
maybe this have other solution.and This problem is from Mathematical olympiad problems
Thank you
 [/INST] Some experimenting reveals that $\pmod{7}$ is the way to go:
$$x^3-5x+10=2^y$$
\begin{align}
\begin{array}{|c|c|}
\hline
x \pmod{7} & x^3-5x+10 \pmod{7} \\
\hline
0 & 3 \\
\hline
1 & 6 \\
\hline
2 & 1 \\
\hline
3 & 1 \\
\hline
4 & 5 \\
\hline
5 & 5 \\
\hline
6 & 0 \\
\hline
\end{array}
&
\begin{array}{|c|c|}
\hline
y \pmod{3} & 2^y \pmod{7} \\
\hline
0 & 1 \\
\hline
1 & 2 \\
\hline
2 & 4 \\
\hline
\end{array}
\end{align}
We see that we must have $3 \mid y$, so let $y=3z$, then 
$$x^3-5x+10=2^{3z}=(2^z)^3$$
This means that the LHS is a perfect cube, which can't happen for large values of $x$ as we may bound it between consecutive perfect cubes. Indeed, for $x \geq 3$, we have 
$$(x-1)^3<x^3-5x+10<x^3$$
(reduces to $0<3x^2-8x+11=x(3x-8)+11$ and $10<5x$ which are true for $x \geq 3$)
Thus there are no solutions for $x\geq 3$. It is now straightforward to check that $x=1$ fails and $x=2$ gives the only positive integer solution $(2, 3)$.
</s>