Prove that $x^{2} \equiv 1 \pmod{2^k}$ has exactly four incongruent solutions 
Prove that $x^{2} \equiv 1 \pmod{2^k}$ has exactly four incongruent solutions.

My attempt:
We have, 
$x^2 - 1 = (x - 1) \times (x + 1)$, then 
$(x - 1)(x + 1) \equiv 0 \pmod{2^k}$
which implies,
$2^k|(x - 1)$ or $2^k|(x + 1) \implies x \equiv \pm 1 \pmod{2^k} (1)$
Furthermore, $2^{k-1} \equiv 0 \pmod{2^k} \Leftrightarrow 2^{k-1} + 1 \equiv 1 \pmod{2^k}$.
Multiply both sides by $-1$, we have another congruent namely $-(2^{k-1} + 1) \equiv -1 \pmod{2^k}$
Hence,  $x \equiv \pm(1 + 2^{k-1}) \pmod{2^k} (2)$
From $(1)$ and $(2)$, we can conclude that $x^{2} \equiv 1 \pmod{2^k}$ have four incongruent solutions.  
Am I in the right track?  
Thanks,

Existence is easy:
The solutions are $\{1,2^{k-1}-1,2^{k-1}+1,2^k-1\}$.
Squaring them gives $\{1, 2^{2k-2}-2^k+1, 2^{2k-2}+2^k+1, 2^{2k}-2^{k+1}+1\}$.
Reducing $\pmod {2^k}$ gives 1 in each case (given that $2k-2 > k+1$, which forces $k \ge 3$).
For uniqueness you could use the fact that the units group of $\mathbb{Z}/2^k \mathbb{Z}$ is $C_2 \times C_{2^{k-2}}$ on the other hand that also gives existence..
