“…Assume (p, d) is one of (2, 4), (2, 10), (2,12), (2,18), so r is 5, 11, 13, 19 respectively. Considering for i all the prime powers dividing 2 d −1 d+1 , which are 3,9,27,5,7,31,73, and taking for k the values 2, 6, 18, 4, 3, 5, 18 respectively (regardless of the value of j), we see that condition (B) fails, so by our observation above i = 1 and G 0 = ΓL(1, 2 d ).…”