Proving that $x(x^2-1)(x^2-10)=c$ cannot have five integer solutions for any real $c$ I found this question that caught my attention at MSE and  I did a solution, but I suspect something is wrong with the solution.

Original problem says:

Prove that for any real values of $c$, the equation $x(x^2-1)(x^2-10)=c$ can't have $5$ integer solutions.


The things I have done:
I find stronger (?) results in this answer:

*

*If $c=0$, then we have $3$ integer solutions.


*If $c>0$ or $c<0$, then we have only $1$ integer solution.
This means,

*

*The number of integer solutions is always less than $4$.


Let,
$$f(x)=x(x^2-1)(x^2-10)$$
and
$$x(x^2-1)(x^2-10)=c$$
where $c\in\mathbb R, x\in\mathbb Z$.
In fact we need only $c\in\mathbb Z$. Because, if $x\in\mathbb Z$ then $c\in\mathbb Z$.
We see that $c=0$ is trivial.
$\underline{\text{Case}-1:~c>0}$
$$\begin{align}&x(x^2-1)(x^2-10)>0 ,x\in\mathbb Z \\ &\iff x\in \left\{-3, -2\right\}∪[4,+\infty)\end{align}$$
Then suppose that, $x_1≥4, x_1\in\mathbb Z $ is a solution.
If $x_2>x_1≥4$ then $f(x)$ is strictly increasing and if $4≤x_2<x_1$, then $f(x)$ is strictly decreasing. This means, if $x_1≥4$, then we have one positive integer solution.
Then, we see that $f(-3)<f(-2)<f(4)$ and $f(x)$ is strictly increasing for $x≥4$.
This implies, if $c>0$, then we have only one integer solution.

$\underline{\text{Case}-2:~c<0}$
Let's multiply both sides of the equation by $(-1).$
$$-x(x^2-1)(x^2-10)=-c, c<0$$
Let, $-x=t$ then
$$t(t^2-1)(t^2-10)=-c,-c>0$$
This means, for $c<0$ we have also one integer solution.
Thus, we conclude the number of integer solutions is always less than $4$.

Does my solution contain any errors?
Please, don't post the correct solution.
Thank you for reviewing.
 A: If $c$ is bigger than the local maximum, $y$ value about
$3.723601932658154682760729230  ,$  then there are, at most, three real solutions to $x(x^2-1)(x^2-10) = c.$
Just draw a horizontal line at $y=c$ for some $c$ value of interest. Then, when $|c| \leq 3.723601932658154682760729230,$  we do get five real roots, but these are not all integers as the largest is strictly between 3 and 4.

A: For your answer my opinion is : your solution is correct because you analyze the problem
in good way.
for the case of $c=0$ it is trivial as you said .
for the cases  $c>0 $ and $c<0$ they are similar so I will discuss with you the case of
$c>0$ , you wrote the intervals of the integer $x$  that satisfy the case  then for
$[4,+\infty)$  you proved that there is only one solution (but the part of your
answer that $f(x)$  is strictly decreasing it is not true and you don't need it because
$f(x)$ is increasing  on $[4,+\infty)$ and this  is enough to prove that we don't have
another roots in this interval ),
and for $x\in \left\{-3, -2\right\}$ ,now if we substitute $x=-3$ in $x(x^2-1)(x^2-10)=c$
we have $c=24$  and for $x=-2$ we have $c=36$ so we can't have $x=-2$ and $x=-3$ as
solution for the same $c>0$ so as you mentioned  there are at most two solutions.
also we can say that we have only one solution for the case $c>0$ because when $c=24$ then
$x=-3$ but we don't have solutions in interval $[4,+\infty)$ because $f(4)=360$ and the
function is increasing on this interval, the same thing for $c=36$
I hope this discussion is helpful for you.
