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
 A: 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 $(2, 3)$.
A: First $(10,4)$ and $(10,6)$ are not solutions, neither is $(10,y)$ for any integer $y$, since $(10)^3-5(10)+10=2^y$ would imply $5\mid2^y$, a contradiction. 
Now, rearrange the given equation to get
$$
(x^2-5)x=2(2^{y-1}-5)\tag{1}
$$
Now either $x=2k+1$ or $x=2k$ for some integer $k>1$; 
let $x=2k+1$, direct substitution gives
$$(4k^2+4k+1-5)(2k+1)=2(2^{y-1}-5)$$
or
$4(k^2+k-1)(2k+1)=2(2^{y-1}-5)$, a contradiction, since it would imply $2\mid5$. 
Hence, $x$ is even and $x^2-5$ is odd (since $d=\gcd(x^2-5,x)\mid5$ so that $d=1$ or $5$, if $d=5$ we get a contradiction from the above equation, hence $d=1$.
CASE 2. $x=2k$, in this case, $k$ is either odd or even, 
i) $k$ is even, hence $x=4t$, for some $t>1$, substitution into the rearranged equation gives $[(4t)^2-5]\cdot4t=2(2^{y-1}-5)$, a contradiction, since it implies $2\mid5$. 
ii) $k$ is odd, $x=4t+2$, substitute:
$[(4t+2)^2-5]*(4t+2)=2(2^{y-1}-5)$ or
$(64t^3+96t^2+28t-2)=2(2^{y-1}-5)$ or
$(64t^2+96t+28)t=2^y-8$ so that $2\mid t$, hence $t=2r$ and $x=8r+2$; 
substitution gives
$[(8r+2)^2-5]\cdot(8r+2)=2(2^{y-1}-5)$ or
$[(8r+2)^2-5]\cdot(4r+1)=(2^{y-1}-5$ or
$64r^3+48r^2+7r+1=2^{y-3}$ 
From here we have $r \equiv 9 \pmod{16}$ or $r=16s+9$ and hence $x=128s+74$, choosing integer values for $s$ can help narrow the search along with $x=8r+2$ where $r$ is odd.
A: This is useful:
Hensel's Lemma:

Let $f(x)$ be a single-variable polynomial with integer coefficients. Let $p^k\neq 1$ be some non-trivial prime power, and suppose that we have an integer $r$ that is a solution to $f(r)\equiv 0\mod p^k$. Then the set of solutions to $f(x)\equiv 0\mod p^{k+1}$ can be computed to be:
  
  
*
  
*If $f'(r)\not\equiv0\mod p$, then the only solution is $x=r+tp^k$ where $t\equiv-(\frac{f(r)}{p^k})f'(r)^{-1}\mod p$ with $0\leq t<p$.
  
*If $f'(r)\equiv0\mod p$ and $f(r)\equiv 0\mod p^k$, then any $x\equiv r\mod p^{k-1}$ (i.e. $x=r+tp^k$ for $0\leq t<p$ ) is a solution.
  
*If $f'(r)\equiv 0\mod p$ and $f(r)\not\equiv0\mod p^k$, then there are no solutions.
  

