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I have a polynomial ${\frac{{{{({z^2})}^p} \pm {p^p}}}{{{z^2} \pm p}}}$ where $p$ is an odd prime number, and I know it splits into two factors $$ \sum_{i = 0}^{p - 1} a_i z^i \text{ and } \sum_{i = 0}^{p - 1} ( - 1)^i a_i z^i $$

For example, when $p=5$ $$ \begin{eqnarray*} \frac{x^{10}-5^5}{x^2-5} &=& x^8+5x^6+25x^4+125x^2+625\\ &=& (x^4 + 5x^3+15x^2+25x+25)(x^4-5x^3+15x^2-25x+25) \end{eqnarray*} $$ Does anyone know a nice method for determining these two factoring polynomials?

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    Does the choice of sign depend on the residue class of $p$ modulo $4$?2012-05-17
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    The factorization exists, iff the sign is equal to $(-1)^{(p-1)/2}$. Given that, one factor is $$\prod_{j=1}^{p-1}\left(x-\left(\frac k p \right)\zeta_p^k\sqrt{p*}\right),$$ where $\zeta_p=e^{2\pi i/p}$, $p^*=(-1)^{(p-1)/2}p$ and $(\frac k p )$ is the Legendre symbol. The other factor is gotten by switching the signs of the zeros. But I guess you knew that, and would like to see a closed formula for the coefficients :-)2012-05-17
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    Do you have any other examples?2012-05-17
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    @lhf, Also sprach Mathematica: For $p=7$ we get $$x^{14}+7^7=\left( 7 + {x^2} \right) \, \left( 343 - 343\,x + 147\,{x^2} - 49\,{x^3} + 21\,{x^4} - 7\,{x^5} + {x^6} \right) \, \left( 343 + 343\,x + 147\,{x^2} + 49\,{x^3} + 21\,{x^4} + 7\,{x^5} + {x^6} \right)$$ and for $p=3$ $$ x^6+3^3=\left( 3 + {x^2} \right) \, \left( 3 - 3\,x + {x^2} \right) \, \left( 3 + 3\,x + {x^2} \right).$$2012-05-17
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    Thanks; this is very helpful. I assume that the product should go from $k=1$ and not $j=1$. I am definitely looking for a closed formula for the coefficients.2012-05-17
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    @BudgieJane, indeed, that was a silly typo. Thinking...2012-05-17
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    @JyrkiLahtonen, thanks. I couldn't get Wolfram Alpha to do this, though now I can :-)2012-05-17
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    I can't help thinking this is related to the Aurifeullian factorizations. See, e.g., http://mathworld.wolfram.com/AurifeuilleanFactorization.html, although I think there's more to Aurifeullian factorization than what's to be found at that site.2012-05-17
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    Sorry, @Gerry, I should have mentioned: yes it is all related to Aurifeuillian factorization and cyclotomic polynomials.2012-05-17
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    The coefficients of those polynomials are mildly complicated exponentials sums. I can't evaluate them at this time (other than Gauss' sums). It is easy to see that (after you strip the factor $\sqrt{p^*}$ from the zeros) that the coefficients are alternately either integers or integer multiples of $\sqrt{p^*}$, but finding a formula takes more time than I can invest on this at this time (and I may not be man enough to do it anyway). Sorry, Jane. Thanks for an interesting question!2012-05-17
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    Oh my gosh; I seem to have found some sort of solution. It involves Lucas's formula for Cyclotomic Polynomials (see Riesel, Table 24).2012-05-17
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    @Jyrki: Thanks for all your help, it is most appreciated.2012-05-17
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    Another example, for $p=11$ is$$161051 + 161051x + 73205x^2 + 14641x^3 -1331x^4 -1331x^5 -121x^6 + 121x^7 + 55x^8 + 11x^9 + x^{10}$$ together with its companion where the signs of the coefficients of the odd powers of x are reversed.2012-05-17
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    If you have found a solution, write it up and post it as an answer! May seem strange, but answering your own questions is explicitly encouraged around here.2012-05-18
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    It's not the sort of solution I was expecting. I was hoping to see something along the lines of extracting square roots by long division (yes, I know this is not a square root, but there are some things in common with square roots), and this solution doesn't do that.2012-05-18

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