# 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.

• Now you see that $c$ has to be an integer at least if this equation has to have $5$ integer solutions May 3 '21 at 16:44
• Try taking the derivative and see how many places the derivative will be zero. The number of roots is at most 1 more than that. May 3 '21 at 16:47
• @Krishnarjun I don't know derivatives, but how many times do we have to take the derivative? If only once, then solving a degree $4$ eqn is tedious as well. Btw, the question is to be done using Descarte's rule of signs, but I'm not able to figure it out.
– user907745
May 3 '21 at 16:59
• @lonestudent I didn't exactly get how your point helps...
– user907745
May 3 '21 at 16:59
• @Crease substitute $u = x^2$ into the derivative and you'll get a quadratic equation May 3 '21 at 16:59

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:

1. $$a+b+c+d+e = 0$$
2. $$ab+ac+ad+ae+...+de = -11$$

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.

• The second sum is not zero. May 3 '21 at 17:17
• i fixed it, should be correct now May 3 '21 at 17:41
• Well done! If it isn't clear, the constant C is 0 since $0$ is always a solution of the equation. May 3 '21 at 17:50

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.

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.

• Wow, this is nice May 5 '21 at 10:54

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.
• @Crease Intermediate Value Theorem, which is in calculus. The idea is a more rigorous expression of Tropsosphere's idea of increasing and decreasing continuous functions. If you're just learning about Descarte in algebra / pre-algebra, just go with Trop's solution. May 3 '21 at 17:45
• My approach arguably does hide some calculus -- at least the only argument for "a quintic polynomial changes direction at most four times" I can think of offhand is to point to its derivative being a quartic. May 3 '21 at 18:10

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 $$4≤x_2, then $$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 $$f(-3) and $$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 $$4$$.

• 1. How can we be sure that if $x$ belongs to integers, then $c$ has to belong to integers too?
– user907745
May 6 '21 at 3:47
• 2. How can we be sure that if $x_1 \ge 4$ or $c>0$ then we have ONLY $1$ +ve integer soln?
– user907745
May 6 '21 at 3:48
• 3. You have done case 2, in a similar way so I didn't get that either. And btw thanks for your answer.
– user907745
May 6 '21 at 3:48

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.

• Most probably that your hint isn't clear enough. IE, If you clarified that "There is at most one solution $\geq \sqrt{10} \Rightarrow \geq 4$ and at most one solution $\leq - \sqrt{10}$, so we just need to check the small cases where $|x | \leq 3$, and require 3 values to be equal, hence .... (FWIW Hint solutions are often a hit or miss. And in cases where a lot of people are typing up solutions, it's often a miss. I prefer Hint solutions though.) May 3 '21 at 17:48