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When considering the polynomials $x^{n^2} \pm (x-1)^{n^2}$ ( $n$ integer > 1 ) i noticed some things that appeared weird to me.

Discriminant($x^{n^2} + (x-1)^{n^2}) = (n^2)^{n^2}$. Discriminant($x^{n^2} - (x-1)^{n^2}) = (n^2)^{n^2 -2}$.

But that was not all , it appears many of the roots of $x^{n^2} \pm (x-1)^{n^2} = 0$ apart from the trivial $\frac{1}{2}$ can be expressed in rootform ( for all $n$ ).

I just did the following but it seems i missed something :

$x^{n^2} \pm (x-1)^{n^2} = 0$

divide by $x^{n^2}$

$1 \pm (1-1/x)^{n^2} = 0$

$(1-1/x) = (\mp 1)^{1/n^2}$

abs formule

But that did not explain the rootforms or discriminant. And it surprised me those zero's could be expressed without the $(\mp 1)^{1/n^2}$ term.

I must say im not an expert in Discriminants or Galois theory. How to explain and prove this ?

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You can write $x^{n^2}$ as $x^{n^2}$ and $\pm$ as $\pm$. For more about typing math at this site see e.g. here, here or here. –  Martin Sleziak Sep 16 '12 at 16:01
    
Thanks Martin Sleziak ! –  mick Sep 16 '12 at 16:11
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1 Answer

To solve the $-$ case (the $+$ case being similar), note indeed that $x\ne0$, then that the equation is equivalent to $(1-1/x)^{a}=1$ for $a=n^2$, hence $1-1/x=\mathrm e^{2\mathrm i\pi k/a}$ for some integer $0\leqslant k\leqslant a-1$. The case $k=0$ being impossible, $x=1/(1-\mathrm e^{2\mathrm i\pi k/a})$ for some integer $1\leqslant k\leqslant a-1$.

Simplifying the fraction yields finally $2x=1+\mathrm i\cot(\pi k/a)$ for some integer $1\leqslant k\leqslant a-1$.

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