“…According to the above theorem, 1, 1, t 2 = t + t 2 , 1, 1, t 3 = t 3 + t 4 , t, 1, t 2 = t, t 4 , 1, t 3 = t 4 , up to the indeterminacy. Tables 3 and 4 in Miller [9] (there is a typo in Table 4: the (4, 0) entry should start at 4 and not 5). The computer found no non-zero solutions of (10), implying that the deleted squares of L(11, 2) and L(11, 3) are not homotopy equivalent.…”