Is Lehmer's polynomial solvable? The degree 10 polynomial
$$\displaystyle x^{10} + x^9 - x^7 - x^6 - x^5 - x^4 - x^3 + x + 1$$
given by D.H. Lehmer in 1933 has the property that its largest real root, $\beta = 1.176280 \cdots$ is the smallest known Salem number. Moreover, it is a folklore conjecture that $\beta$ is in fact the smallest Salem number.
However, it is curious that one cannot find a reference for the explicit value of $\beta$. I suspect that this is because Lehmer's polynomial is not solvable. Is this the case? If so, is there a reference/relatively simple argument? Furthermore, if Lehmer's polynomial is in fact not solvable, then what is its Galois group?
Thanks for your time.
 A: Lehmer's polynomial is symmetrical,
so $x + x^{-1} =: y$ satisfies a polynomial of half the degree.
It turns out that this is the quintic $y^5 + y^4 - 5y^3 - 5y^2 + 4y + 3 = 0$, 
whose Galois group is the unsolvable $S_5$ (for instance, it's irreducible
$\bmod 2$ and decomposes as $(y^2-2y-1)(y^3-2y^2+2y+2)$ $\bmod 5$,
so the Galois group is a subgroup of $S_5$ of order divisible by $30$ that contains an odd permutation,
and the only such subgroup is $S_5$ itself).  Hence Lehmer's polynomial
is not solvable either.
[It turns out that $y$ generates the totally real quintic field
of third-smaller discriminant $36497$.
By the way, even if a polynomial is solvable, exhibiting a
solution in radicals may not be of much use; for instance,
the Salem root of $x^8 - x^5 - x^4 - x^3 + 1 = 0$ satisfies
$x + x^{-1} = y$ where $y$ is a solution of the quartic
$y^4 - 4y^2 - y + 1 = 0$ with Galois group $S_4$,
but even though this group (and thus also the octic
$x^8 - x^5 - x^4 - x^3 + 1$) is solvable I doubt that you
really want to ponder the explicit formula for $x$
involving things like the cube roots of $187/54 \pm \sqrt{-1957/108}$ . . .]
