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Let $f(x)$ be an irreducible polynomial in $\Bbb{Q}[x]$. I was trying to study these polynomials, and I notice a peculiar symmetry amongst the roots, although I'm not always sure what kind of symmetry the roots have. For instance, the polynomial $x^2-2$ has the roots $\pm\sqrt{2}$. Clearly, the roots show reflection symmetry about the $y$-axis. The roots of $x^2-n$ show a similar symmetry for any positive integer $n$ which is not a perfect square (consider $x^2-3$ for example). Now consider the polynomial $x^4-10x^2+1$. The roots it has are $\pm\sqrt{2}\pm\sqrt{3}$. Clearly, the pairs of roots $\sqrt{2}+\sqrt{3}, -(\sqrt{2}+\sqrt{3})$ and $\sqrt{2}-\sqrt{3},-(\sqrt{2}-\sqrt{3})$ show reflection symmetry about the $y$-axis. However, the fact that we can obtain all roots of this polynomial by just changing the signs in front of the radicals cannot just be a co-incidence. I imagine that there is a symmetry between all the roots, and not just between distinct pairs of them.

Would someone like to comment on this? What kind of symmetry would that be?

EDIT: If all irreducible polynomials over $\Bbb{Q}[x]$ do not have such symmetric roots, could someone comment on what subclass of such polynomials have symmetric roots?

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Writing $X^2-c=(X-\alpha)(X-\beta)$, where $\alpha$ and $\beta$ are the roots of $X^2-c$, shows that the sum of the roots equals $0$. This gives the symmetry you found. Similarly $$x^4-10x^2+1=(x^2-(\sqrt{2}+\sqrt{3})^2)(x^2+(\sqrt{2}+\sqrt{3})^2),$$ is the product of such two polynomials, again giving the symmetry you found.

The study of the symmetries of solutions to polynomial equation is the origin of Galois theory. Modern treatments require some background in group, ring and field theory, but the beauty of Galois theory is well worth the effort of learning basic abstract algebra, in my humble opinion.

Servaes
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