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I have a polynomial of the form:

$$\sum_{m=0}^k\frac{(-1)^{m+1}(4k-2m)!x^{2k-2m}}{m!(2k-m)!(2k-2m+1)!}$$ or identically:

$$\sum_{m=0}^k\frac{(-1)^{m+1}(4k-2m)!(x^{2})^{k-m}}{m!(2k-m)!(2k-2m+1)!}$$

where we can see that the sum of the coefficients is always zero.

Now , my question is that :

what can be concluded from this (i.e sum of the coefficients being zero) about zeros (roots) of the polynomial?

edit:

except the fact that 1 is a root.

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The sum of the coefficients is zero if and only if $1$ is a root of the given polynomial.

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  • $\begingroup$ right, but I mean more than that. (Indeed , $x$ here is $cos(\alpha)$ and other solutions( other than 1 and 0) are of interest to me. $\endgroup$ – Mo_ Mar 10 '13 at 18:16
  • $\begingroup$ from the fact that sum of coefficient is 0 you may conclude only that because of "if and only if" $\endgroup$ – RiaD Nov 16 '17 at 13:53

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