# How to show that a root of the equation $x (x+1)(x+2) … (x+2009) = c$ can have multiplicity at most 2?

How to show that a root of the equation $$x (x+1)(x+2) ....... (x+2009) = c$$ can have multiplicity at most 2 , and to find the value of $c$ for which this is possible.

I proceeded by using the derivative method, but the number of terms involved is huge.I have been trying to find some common property for all these terms, but I've come up with nothing so far.

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If you try it first with degrees 3 and 4 (possibly 5) and see if you see some pattern emerging, that might be the way to go. I sincerely doubt there is anything special about 2009 in this problem (only that the problem itself was written four years ago). –  Arthur Mar 18 '13 at 10:43

Let denote by $$P(x)=x (x+1)(x+2) ....... (x+2009) - c .$$

We have $$P(0)=P(-1)=\cdots P(-2009)=-c$$ and by Rolle's theorem there's $$0>t_0>-1>\cdots>t_{2008}>-2009$$ s.t. $$P'(t_0)=P'(t_1)=\cdots=P'(t_{2008})=0.$$ Since the polynomial $P$ has the degree $2010$ then the degree of $P'$ is $2009$ and then the $t_i$ are the roots of $P'$.

If $c=0$ then the roots of $P$ are different from the roots of $P'$ and there is not a multiple root.

If $c\neq 0$, we apply the Rolle's theoem on $P'$ and we find the roots $s_i$ of $P''$ s.t $$t_0>s_0>t_1>\cdots>s_{2007}>t_{2008}$$ and if we have $P(\alpha)=P'(\alpha)=0$ then $\alpha=t_{i_0}$ ($\alpha$ is a root of $P'$) and then $\alpha\neq s_i\forall i=0,\ldots,2007$ and hence $P''(\alpha)\neq 0$.

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Same as above, thought I should include more for clarity on finding $c$. Let $f(x) = x(x+1)(x+2) \dots (x+2009)$ and $g(x) = f(x) - c$. Clearly $f$ has $2010$ distinct roots, so $f' = g'$ has $2009$ distinct roots as well. However if $g$ has a root of multiplicity more than $2$, that root must have multiplicity in $g'$ as well. Hence $g$ cannot have any root of multiplicity more than $2$. For finding $c$, let $\alpha$ be any of the 2009 roots of $f'(x)$. Then if you let $c = f(\alpha)$, we will have $g(\alpha) = f(\alpha) - c = 0$ and $g'(\alpha) = f'(\alpha) = 0$ by choice. –  Macavity Mar 18 '13 at 11:58
@Macavity I too immediately thought $P'$ has simple roots, so $P$ cannot have triple roots. If your comment had been an answer, I would have voted for it. –  Marc van Leeuwen Mar 18 '13 at 12:28
Nice answer +1... –  DonAntonio Mar 18 '13 at 12:47
@Macavity, your comment really cleared things up for me.I was a bit stuck with Sami Ben's proof, especially with the Rolle's Theorem part.However you comment did the job. So I'm gonna mark this answer as accepted. –  donvoldy666 Mar 18 '13 at 14:44

For a more pedestrian approach involving explicit derivative computations, observe that if $c=0,$ then all roots have multiplicity $1.$ So we only consider $c\ne0.$ Let $f(x)=x(x+1)(x+2)\ldots(x+2009)$ and $g(x)=f(x)-c.$ There is a root with multiplicity greater than $2$ only if there is an $\alpha$ such that $g(\alpha)=g'(\alpha)=g''(\alpha)=0.$ But $$g'(x)=\frac{f(x)}{x}+\frac{f(x)}{x+1}+\frac{f(x)}{x+2}+\ldots\frac{f(x)}{x+2009}=\sum_{j=0}^{2009}\frac{f(x)}{x+j}$$ and $$g''(x)=\sum_{0\le j<k\le2009}\frac{2f(x)}{(x+j)(x+k)}=\frac{[g'(x)]^2}{f(x)}-\sum_{j=0}^{2009}\frac{f(x)}{(x+j)^2}.$$

If $g(\alpha)=g'(\alpha)=g''(\alpha)=0,$ then $$f(\alpha)=c,\qquad g'(\alpha)=\sum_{j=0}^{2009}\frac{c}{\alpha+j}=0,\qquad g''(\alpha)=-\sum_{j=0}^{2009}\frac{c}{(\alpha+j)^2}=0.$$ But this is impossible since all terms in the expression for $g''(\alpha)$ are non-zero of the same sign.

We have double roots for solutions $\alpha$ of the equation $$\sum_{j=0}^{2009}\frac{1}{\alpha+j}=0,$$ and the values of $c$ for which such a double root occurs are $f(\alpha).$ Note that if $\alpha$ is a double root, then so is $-2009-\alpha,$ and that the corresponding values of $c$ are equal. Hence those $c$ that give rise to double roots actually give rise to pairs of double roots.

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