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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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    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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Here is the solution I found, which to me is unsatisfactory.

This solution uses Table 24 from Riesel’s “Prime Numbers and Computer Methods for Factorization”. Look up the number you are interested in in the n column, and note down the two sets of coefficients given there: they are $U_n(x)$ and $V_n(x)$. Multiply each coefficient in $U_n(x)$ by $p^0$, $p^1$, $p^2$, …, $p^{p-1}$, and those in $V_n(x)$ by $p^1$, $p^2$, … $p^{p-1}$. Then take a coefficient from each list alternately and you have your factor. Do the same for the other factor after multiplying $V_n(x)$ by $-1$.

Example for $p=11$:
$U_n(x)=1, 5, -1, -1, 5, 1$
$V_n(x)=1, 1, -1, 1, 1$
Multiplying by powers of $11$:
$U_n = 1, 55, -121, -1331, 73205, 161051\\ V_n = 11, 121, -1331, 14641, 161051$
Taking the coefficients one at a time from each list one factor is $x^{10} + 11x^9 +55x^8 +121x^7 -121x^6 -1331x^5 -1331x^4 +14641x^3 +73205x^2 +161051x +161051$ and the other factor is

$x^{10} - 11x^9 +55x^8 -121x^7 -121x^6 +1331x^5 -1331x^4 -14641x^3 +73205x^2 -161051x +161051$

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    I think I could prove it if I put my mind to it. Howeveer, I'm not too bothered, as this is not the solution I'm looking for. (Maybe what I'm looking for is unobtainable). Your second comment here mirrors what's in Paul Garrett's paper. See my comments to the next solution.2012-05-18
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Here is a follow-up from my comment, but unfortunately it does not work. Since it is too long for a comment, I leave it here as an answer.

Set $z=\lambda x$, where $\lambda= \sqrt{p}$ for convenience. Then your polynomials are $p^{p-1}\dfrac{(x^2)^p\pm1}{x^2\pm1}$.

Now $(x^2)^p-1=x^{2p}-1=\Phi_{2p}(x)\Phi_p(x)\Phi_2(x)\Phi_1(x)=\Phi_p(-x)\Phi_p(x)(x^2-1)$, where $\Phi_n$ is the $n$-th cyclotomic polynomial.

Then $p^{p-1}\dfrac{((x^2)^p-1)}{x^2-1}=\lambda^{2p-2}\Phi_p(-x)\Phi_p(x)=(\lambda^{p-1}\Phi_p(-x))(\lambda^{p-1}\Phi_p(x))=Q(-z)Q(z)$, where $Q(z)=\lambda^{p-1}\Phi_p(x)=\lambda^{p-1}\Phi_p(z/\lambda)$.

This is the correct form from the original observations. Unfortunately, $Q$ does not have integer coefficients. The factorization obtained for the case $p=5$ does not come from this one. And there goes my idea...

Edit: I've now found that all this is discussed on page 6 of these notes by Paul Garrett. I think this is the best answer to the question.

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    @BudgieJane, you might have mentioned Garrett's paper...2012-05-19