How does there exist $2d$ solutions in the second part of the above proposition?
My process is following: I know that $\{ (-1)^a 5^b | a=0,1 $ and $ 0 \leq b < 2^{e-2} \}$ is a reduced residue system $mod $ $2^e$.
Let $n $ be even. Let $a \equiv (-1)^s 5^t (2^e) $.
Let a solution $x$ exists. Let the solution be $(-1)^y 5^z (2^e)$.
Then $x^n \equiv a(2^e)$ $\iff$ $(-1)^{yn} 5^{nz} \equiv (-1)^s 5^t (2^e) \iff 5^{nz} \equiv (-1)^s 5^t (2^e)$.
Now $e \geq 3 \Rightarrow 5^{nz} \equiv (-1)^s 5^t (4) \iff 1 \equiv (-1)^s (4) \iff s=0$.
So we get $5^{nz} \equiv 5^t (2^e) \iff nz \equiv t (2^{e-2})$ since the order of $5 $ mod $2^e$ is $2^{e-2}$.
Since $nz \equiv t (2^{e-2})$ has a solution iff $d = gcd(n, e^{e-2})| t$ and there are exactly $d$ solutions.
I am confused as to how there are $2d$ solutions.