10. Show that the congruence $x^{2} \equiv 1\left(\bmod 2^{k}\right)$ has exactly four incongruent solutions, namely $x \equiv \pm 1$ or $\pm\left(1+2^{k-1}\right)\left(\bmod 2^{k}\right)$, when $k>2$. Show that when $k=1$ there is one solution and when $k=2$ there are two incongruent solutions.