Solution to Problem 4

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Correct Answer:

There exists no such integer \(r\) satisfying the given conditions.

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Correct answers were given by Manjunatha M R (Infosys) and Nagendra P (Research Scholar, UOM). However, none gave complete justification.

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Solution.

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Suppose \(q\) was the repeated root of \(p(x)\). Then, we must have

\(p(x) = (x-q)^2(x-a)\),

where \(a\) is the third root of \(p(x)\). It is very easy to see that \(q\) must also be a root of the derivative of \(p(x)\). That is, \(q\) is a root of

\(p'(x) = 3x^2-2(r-1)x-11\).

That is, we must have that

\(3q^2-2(r-1)q-11 =0\),

which implies that

\(q= \dfrac{2(r-1)±\sqrt{4(r-1)^2+132}}{6} = \dfrac{(r-1)\pm\sqrt{(r-1)^2+33}}{6}\).

As \(q\) is an integer, the term inside the squer root must be a perfect square. That is

\((r-1)^2+33 = b^2\),

for some integer \(b\). The above equation implies that

\(33 = b^2-(r-1)^2 = (b-r+1)(b+r-1)\).

Now, by fundamental theorem of arithmetic, we have the following choices for \(b-r+1\) and \(b+r-1\):

Case	\(b-r+1\)	\(b+r-1\)	\(r-1\)	Possible \(q\)
1	\(\pm 1\)	\(\pm 33\)	\(\pm 16\)	\(\pm 11\)
2	\(\pm 3\)	\(\pm 11\)	\(\pm 4\)	\(\mp 1\)
3	\(\pm 11\)	\(\pm 3\)	\(\mp 4\)	\(\pm 1\)
3	\(\pm 33\)	\(\pm 1\)	\(\mp 16\)	\(\mp 11\)

Note that, to construct the above table, we have used the condition that \((b-r+1)(b+r-1)=33\). Now, if we use the value of \(r\) and the corresponding value of \(q\) from each case, we can easily see that for no such \(r\) and \(q\), the polynomial \(p(x)\) will have \(q\) as its root. To illustrate what we are telling, we take an example. Readers can verify the same for others.

In the case 2, when \(b-r+1=-3\) and \(b+r-1=-11\), we have \(r-1=-4\) (i.e. \(r=-3\)) and the possible \(q\) is -1. Using \(r=-3\) in \(p(x)\), we get the polynomial

\(x^3+4x^2-11x-12\).

Clearly, by direct inspection it can be easily seen that \(q=-1\) is not a root of the above polynomial. Hence, the contradiction forces us to the conclusion that there can exist no such \(r\) satisfying the given conditions.

Note. We were in fact surprised after making an observation (which was mainly due to Manjunatha M R) that the polynomial \(x^3-(r-1)x^2-11x+4r\) can never have two real repeated roots for any integer \(r\).