Find Possible values of $c$ such that polynomial $$x(x+1)(x+2) \cdots (x+2009)=c$$ has one root of multiplicity two
My Try:
Assuming the polynomial equation has root $k$ Repeated twice The given polynomial can be written as
$$f(x)=x(x+1)(x+2)\cdots (x+2009)-c =(x+k)^2 \left(x^{2008}+a_1x^{2007}+a_2x^{2006}+\cdots-\frac{c}{k^2}\right)$$
Now $f'(x)=0$ should have $k$ as a root with multiplicity One
we have
$$f'(x)=f(x)\left(\frac{1}{x}+\frac{1}{x+1}+\frac{1}{x+2}+\cdots+\frac{1}{x+2009}\right)$$
can we use this concept to find $c$
A plot of $x(x+1)(x+2)(x+3)$ is below. Your polynomial will have lots more squiggles but be similar. You get a double root any time $c$ matches one of the local maxima or minima. You should be able to convince yourself that your polynomial is symmetric around $x=-1004.5$. If you want exactly one double root, that will only happen when $c$ matches the central peak. The value there is $$\prod_{n=0}^{1004}\left(n+\frac 12\right)^2 \approx 5.28\cdot 10^{5161}$$If you want at least one double root there will be $1004$ more values, which I think will be hard to find.