We call $\alpha(z) = a_0 + a_1 z + \dots + a_{n-1} z^{n-1}$ a Littlewood polynomial if $a_j = \pm 1$ for all $j$. We call $\alpha(z)$ self-reciprocal if $\alpha(z) = z^{n-1}\alpha(1/z)$, and call $\alpha(z)$ skewsymmetric if $n = 2m+1$ and $a_{m+j} = (-1)^j a_{m-j}$ for all $j$. It has been observed that Littlewood polynomials with particularly high minimum modulus on the unit circle in $\bC$ tend to be skewsymmetric. In this paper, we prove that a skewsymmetric Littlewood polynomial cannot have any zeros on the unit circle, as well as providing a new proof of the known result that a self-reciprocal Littlewood polynomial must have a zero on the unit circle.