Show that the congruence x 2 ≡ 1 (mod 2k ) has exactly four incongruent solutions, namely x ≡ ±1 or ± (1+2k−1 ) (mod 2k ), when k > 2. show that when k = 1 there is one solution and when k = 2 there are two incongruent solutions. (a) solve x 2 ≡ 1 (mod 2)