How find this $x^3-5x+10=2^y$

Question

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

Answer 1

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 $(2, 3)$.

Answer 2

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.

Answer 3

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:

1. 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 $0\leq t<p$.
2. 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.
3. If $f'(r)\equiv 0\mod p$ and $f(r)\not\equiv0\mod p^k$, then there are no solutions.