Three, (50 points) Solve the equation $x^{2} \equiv 1\left(\bmod 2^{t}\right)$.


When $t=1$, the number of solutions is $1, x \equiv 1(\bmod 2)$. When $t=2$, the number of solutions is $2, x \equiv \pm 1\left(\bmod 2^{2}\right)$.
When $t \geqslant 3$, write the congruence equation as: $(x-1)(x+1) \equiv 0\left(\bmod 2^{t}\right)$. Since $x$ must be odd, let $x=2 m+1$, substituting gives $4 m(m+1) \equiv 0\left(\bmod 2^{t}\right)$, which simplifies to $m(m+1) \equiv 0\left(\bmod 2^{t-2}\right)$.
Therefore, it must be that $m \equiv 0,-1\left(\bmod 2^{t-2}\right)$, so $x \equiv 1,-1\left(\bmod 2^{t-1}\right)$,
Thus, $x \equiv 1,1+2^{t-1},-1,-1+2^{t-1}\left(\bmod 2^{t}\right)$.