There is no square-complementary graph of girth 6

A graph is {\it square-complementary} ({\it squco}, for short) if its square and complement are isomorphic. We prove that there is no squco graph of girth $6$, thus answersing a question asked by Milani\vc et al. [Discrete Math., 2014, to appear], and leaving $g = 5$ as the only possible value of $g$ for which the existence of a squco graph of girth $g$ is unknown.Comment: 4 pages, 1 figur