A blog for enthusiastic math-lovers!

Another Summation Problem

Please continue to work on the previous problem; I will post hints or solutions to that in the near future!

Regardless: A cool pattern of polynomials goes as follows-

(x+1)^2=x^2+2x+1

(x^2+x+1)^2=x^4+2x^3+3x^2+2x+1

(x^3+x^2+x+1)^2=x^6+2x^5+3x^4+4x^3+3x^2+2x+1

….

See if you can use this little tip to simplify the following problem (which comes from AIME 2003, number 15!). It combines this tip and the ideas emphasized in the previous post!

Let P(x) = 24x^{24}+\sum_{j = 1}^{23}(24-j)(x^{24-j}+x^{24+j}).

Let z_{1},z_{2},\ldots,z_{r} be the distinct zeros of P(x), and letz_{k}^{2}= a_{k}+b_{k}i fork = 1,2,\ldots,r, where i =\sqrt{-1}, and a_{k} and b_{k} are real numbers.

Let \sum_{k = 1}^{r}|b_{k}| = m+n\sqrt{p}, where n, m  and p are integers and p is not divisible by the square of any prime. Find n+m+p.

Note that this problem involves multiple steps; even if you don’t know the entire solution to it, feel free to comment with intermediate steps or ideas you may have.

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