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

My teacher gave this question at the end of the class as a "challenging" one. I have no idea how to approach it. He also gave a hint that opening the brackets might help, so I did so and got the following:
$$ x^5-11x^3+10x-c=0$$
I don't see how this is helping, and finding roots of this equation in terms of $c$ is just tedious. Any hints or help would be most appreciated!

The question is to be done using Descarte's rule of signs. Other methods are welcome too.
 A: Here's a more "elementary" way.
Let a,b,c,d,e be integer roots. We attempt to find a contradiction.
By Vieta's formulas, we have:

*

*$a+b+c+d+e = 0$

*$ab+ac+ad+ae+...+de = -11$
So
$22 = (\text{equation}_1)^{2} - 2\cdot(\text{equation}_2) $
$= a^{2}+b^{2}+c^{2}+d^{2}+e^{2}$
Note that $a^{2},b^{2},c^{2},d^{2},e^{2}$ must each be one of 0,1,4,9,16.
One can quickly check (there are not that many cases) that, up to permutation, only $(a^2,b^2,c^2,d^2,e^2) = (16,4,1,1,0) \text{ or } (9,4,4,4,1,0)$ work.
This implies that the constant of our polynomial "C" (capital) is 0.
Then the polynomial becomes $x(x^2-1)(x^2-10)=0$ which clearly doesn't have 5 integer roots.
A: Let $f(x) = x(x^2-1)(x^2-10)$.
Since $f$ has roots $\{0,\pm 1,\pm\sqrt{10}\}$, the monotonicity of $f$ changes from decreasing to increasing somewhere between $-1$ and $0$, and back from increasing to decreasing somewhere between $0$ and $1$.
If $f(x)=c$ has five real solutions, one of those solutions has to lie in that increasing interval of $f$ (we know a degree-5 polynomial changes direction at most 4 times, so for 5 crossings with a horizontal level each of the monotonicity intervals needs to contribute one). The only integer in this interval is $0$, so if all five solutions are integers, $0$ must be one of them. But then $c=f(0)=0$.
But the five roots of $f(x)=c$ are not all integers.
A: Since $c$ having five integer solutions means $-c$ does as well, and $c = 0$ obviously doesn't work, we can assume $c > 0$ without loss of generality.
Applying Descartes' rule of signs tells us that there are at most three positive solutions and at most two negative solutions. So if there are five real solutions, it must be that three are positive and two are negative. Now, consider the polynomial obtained by replacing $x$ by $x+1$:
$$
x^5 + 5x^4 - x^3 - 23x^2 -18 x - c.
$$
Descartes rule of signs says there is at most one positive solution and at most four negative ones. So, by shifting the polynomial by 1, we have moved two roots from positive to negative. Thus, those roots were in the interval $(0, 1)$, and could not have been integers.
A: Descarte alone (blindly) isn't sufficient, since there could be 5 real roots (like when $ c = 0$).
Let $f(x) = x (x^2 - 1) (x^2 - 10)$ .
Suppose that $f(x) = c$ has 5 integer solutions.
Notice that $f(x) = 0 $ has the 5 real solutions $ - \sqrt{10}, - 1, 0, 1, \sqrt{10}$, and these are all of them.
By Intermediate Value Theorem / considering the shape of the graph, if $f(x) = c$ has 5 real solutions, then there must either be:

*

*Case 1: 1 solution in $( - \infty,  -\sqrt{10} ] $, 2 solutions in $ [ -1, 0 ] $, 2  solutions in $[1, \sqrt{10} ] $, or

*Case 2: 2 solutions in $ [ -\sqrt{10} , -1 ] $, 2  solutions in $[0, 1 ] $, 1 solution in $[ \sqrt{10} , \infty) $,

Case 1 (resp 2): Since there are only 2 integers in $ [-1, 0]$ (resp $[0,1]$) and $f(0) = f(1) = 0$, we conclude that $c = 0 $ (resp 0). But $f(x) = 0 $ doesn't have 5 integer solutions.

Notes

*

*This generalizes to cases of $ f(x) = x (x^2 -1 ) (x^2 - d)$ where $ d $ is not a perfect square.

*The general case, $ f(x) = x \prod (x^2 - d_i) $, where not all of the $d_i$ are perfect squares, seems interesting to me. We could proceed in a similar manner in some cases.

*

*(Idea of Troposphere / Calvin / Eyeballfrog) If there exists a $n^2 <  d_i  , d_j < (n+2)^2$ (and slight variations), then there are no solutions .

*(Idea of Kyary, Lone Student, Improve) Check values till $N^2 > \max d_i$, seeing if enough of them are equal.



A: We'll get 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 if $x\in\left\{-3,-2\right\}$, then $f(-3)<f(-2)<f(4)$ 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.
Finally, we conclude that the number of integer solutions is always less than $4$.
A: This method does not involve the Descartes' rule of signs.
Hint:
Consider the function $f(n) = n(n^2-1)(n^2-10)$.
There exists a small integer $N$ such that $f(n+1)>f(n)$ for all $n$ larger than $N$ and $f(n+1)>f(n)$ for all $n$ smaller than $-N$. So if there are $5$ integers $n_1$, $n_2$, $n_3$, $n_4$ and $n_5$ such that $f(n_1) = f(n_2) = f(n_3) = f(n_4) = f(n_5) = c$，then most of them have to be quite small (in absolute value).
That is, you don't have to go far away from the origin before $f$ grows so quickly that no values are attained by more than one $n$. So you only have to check a few integer values  really close to the origin.
