Find sum of all integral values of $r$ such that all roots of the equation $x^3-(r-1)x^2-11x+4r=0$ are also integers

Question

Find sum of all integral values of $r$ such that all roots of the equation $$x^3-(r-1)x^2-11x+4r=0$$ are also integers.

What I could do was $$r=\frac{x^3+x^2-11x}{x^2-4}=x+1+\frac{4-7x}{x^2-4}$$ Since $r$ and $x$ are both integers $(4-7x)$ must be divisible by $(x^2-4)$

After this I couldn't proceed. There could be a way using Viete's formulae but that also did not lead to solution

Answer 1

Let $p(x)=x^3-(r-1)x^2-11x+4r$ then $p(2) = -10 \lt 0$ so the cubic has a root $\gt 2$ or, being an integer root, $\ge 3$.

For $r=x+\frac{x^2-7x}{x^2-4}$ to be an integer, $|x^2-4| \le |x^2-7x|$ $\iff 0 \le (x^2-7x)^2 - (x^2 - 4)^2$ $=-(2 x + 1)(7x - 4) (x - 4)$ $\iff x \le - \frac{1}{2} \;\lor\; \frac{4}{7} \le x \le 4$. It follows that the root $\ge 3$ can only be $3$ or $4$. It cannot be $3$ because $3^2 - 4 \not\mid 3^2 - 7 \cdot 3$, which leaves only $\,x=4\,$ to check.

Answer 2

The previous answer has been accepted by the poser. Yet, posting this answer as it is entirely different from the previous one.

Set $p(x) = x^3-(r-1)x^2-11x+4r$. We know that if $q$ is an integral root of $p(x)$, then $q$ must be a factor of $4r$. Let $4r=qk$. Hence, we must have $$q^3-\left(\dfrac{qk}{4}-1\right)q^2-11q+qk=0.$$ Simplifying, we see that $$(4-k)q^2+4q+(4k-44)=0.$$ Note that $q$ is an integer. Also, notice that $$q=\frac{-4\pm\sqrt{16-4(4-k)(4k-44)}}{2(4-k)}=\frac{-2\pm2\sqrt{k^2-15k+45}}{(4-k)}.$$ Hence, we must have that $k^2-15k+45$ must be a perfect square. That is $$k^2-15k+45 = \left(k-\frac{15}{2}\right)^2-\frac{45}{4}=a^2.$$ Or equivalently $$(2k-15)^2-b^2=45$$ where $b=2a$. In other word $$(2k-15-b)(2k-15+b)=45.$$ Thus $(2k-15-b)$ is a factor of $45$. Hence the possible values for the same are $$\pm1, \pm3, \pm5, \pm9, \pm15, \pm45.$$ Correspondingly, $(2k-15+b)$ has the values respectively: $$\pm45, \pm15, \pm9, \pm5, \pm3, \pm1.$$

We can now consider different cases and see what may happen. This has been summarized in the table below:

$$\begin{array}{|c|c|c|c|c|c|} \hline Sl. No. & 2k-15-b & 2k-15+b & k & q & \text{Remark}\\ \hline 1 & \pm 1 & \pm 45 & k=19, -4 & -3 \,\, \text{when $k=-4$} & r=3, \, \text{$p(x)=0$ }\\ & & & & \text{$k=19$ doesn't yield integral $q$} & \text{has three integral roots}\\ \hline 2 & \pm 3 & \pm 15 & k=3, 12 & 4 \,\, \text{when $k=3$} & r=3, \,\, \text{when $k=3$ and $p(x)=0$ has three integral roots}\\ & & & & \text{1 when k=12} & r=3, \,\, \text{when $k=12$ and $p(x)=0$ has three integral roots}\\ \hline 3 & \pm 5 & \pm 9 & 4, 11 & 0 & r=0,\,\, \text{$p(x)=0$ doesn't have }\\ & & & & \text{$k=4$ is not allowed} & \text{all integral roots}\\ \hline \end{array} $$ We need not verify other cases for $2k-15-b$ (i.e. $2k-15-b = \pm9, \pm15, \pm45$) as they are the repetition of the above cases. Thus: $r=3$ is the only possibility in which case the three roots clearly are $q=-3, 1, 4$ as seen from the above table.