Show that $x^2\equiv a \pmod {2^n}$ has a solution where $a\equiv1 \pmod 8$ and $n\ge3$

Show that $$x^2\equiv a \pmod {2^n}$$ has a solution where $a\equiv1 \pmod 8$ and $n\ge3$

Actually, the question I had to solve was more complicated something like this:

$x^2\equiv a \pmod{2^n}$ has a solution where $n \ge3 $ iff $a\equiv1 \pmod 8$ and the equation has exactly 4 incongruent solutions.

I figured out every other things. First, when the equation do have a solution, then $a\equiv1 \pmod{8}$ and the equation has exactly 4 incongruent solutions. The problem is that I couldn’t prove that the equation do have at least one solution when $a\equiv1 \pmod{8}$, which is the reverse part of the proof. That is the only remaining part of my proof.

Could anybody fill in the missing part of my proof? Thanks in advance.

(The hard part of this problem for me was that I couldn’t use Legendre symbols or primitive roots something like that since it is about $\pmod {2^n}$.)

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