Quadratic congruence There is question

For each of prime $p$, show that the congruence 
  $x^2 \equiv1 \pmod {p^a}$
  has precisely two solutions. 
Continue and show that the congruence
  $x^2 \equiv 1 \pmod {2^a} $
  has one solution if $a=1$, two solutions if $a=2$, and four solutions if $a \ge 3$.

I don't know how to do. Help please?

A hint has been given already for odd primes. So let us deal with $2^a$. If $a=1$ or $a=2$, you should be able to verify the assertion, by just calculating. 
For practice, we can also do an explicit computation for $a=3$. It is easy to verify that $1^2\equiv 1\pmod{8}$, that $3^2\equiv 1\pmod{8}$, that $5^2\equiv 1\pmod{8}$, and that $7^2\equiv 1\pmod{8}$. And of course if $x$ is even then we cannot have $x^2\equiv 1\pmod{8}$, so there are $4$ solutions modulo $8$. 
Now let us look at general $a\ge 3$. Suppose that $x^2\equiv 1\pmod{2^a}$. This can be rewritten as 
$$(x-1)(x+1)\equiv 1\pmod{2^a}.$$
Note that $x$ must be odd, so both $x-1$ and $x+1$ are even. If $x-1$ is congruent to $0\pmod{4}$, then $x+1\equiv 2\pmod{4}$. And if $x-1\equiv 2\pmod{4}$, then $x+1\equiv 0\pmod{4}$. 
So if $x$ is odd, one of $x-1$ or $x+1$ is divisible by $4$, and the other is divisible by $2$ but by no higher power of $2$. 
If $2^a$ divides $(x-1)(x+1)$, where $a\ge 3$, there are $4$ possibilities: 
(i) $2^a$ divides $x-1$, that is, $x\equiv 1\pmod{2^a}$. Informally, $x-1$ all by itself contributes enough $2$'s. 
(ii) $2^a$ divides $x+1$, that is, $x\equiv -1\pmod{2^a}$. Informally, $x+1$ contributes enough $2$'s.
(iii) $2^{a-1}$ divides $x-1$, but $2^a$ doesn't. Informally, $x-1$ does not quite have enough $2$'s, but $x+1$ chips in with the only $2$ it has. Then $x-1\equiv 2^{a-1}\pmod{2^a}$, that is, $x\equiv 1+2^{a-1}\pmod{2^a}$. 
(iv) $2^{a-1}$ divides $x+1$, but $2^a$ doesn't. That gives $x\equiv -1+2^{a-1}\pmod{2^a}$.  
It is almost obvious that if $a\ge 3$, these $4$ solutions are incongruent modulo $2^a$.
