I was encountered one example in Elementary Number theory By Jones and Jones .
I understand whole example except what is argument for number of soultion congruence classes for $n=2 \mod 4$ as $2^{k-1}$. I do not understand .
Any Help will be appreciated
Let $n=p_1^{e_1}\dotsm p_k^{e_k}$. Now if $n$ is odd it is made up of odd prime powers in its factorisation. We know if $p$ is an odd prime there are just two classes of solutions of $x^2\equiv 1\pmod{p^{e}}$, $x\equiv\pm1$. This gives, for each odd prime $p_i$, $N_i=2$ classes in $\mathbb{Z}_{p_i}$, and so $N=2^k$ in this case.
If $p=2$ then $1\equiv-1$ modulo $2$, so there is just one class: $x\equiv1$.
If $p=4$ there are two classes $x\equiv2^{2-1}\pm1\equiv\mp1$. If $p=2^e>8$ there are four given by $x\equiv\pm1$ and $x\equiv2^{e-1}\pm1$, that is, as well as $1$ and $-1\equiv7$, we also get $3$ and $5$ as solutions to $x^2\equiv 1\pmod{2^{3}}$, with this pattern continuing for $e\ge3$.
Now as we add and subtract $1$ from an odd number, on looking at the factorisation $$x^2-1=(x+1)(x-1)$$ we see both factors are even, with one factor divisible by $4$ and the other not, i.e., one must be $\equiv2\pmod{4}$. In fact the one divisible by $4$ is divisible by $2^{e-1}$.
Now since there are $N=N_1\dotsm N_k$ classes in $\mathbb{Z}_n$, and since if $p_1=2$, then $N_1=1$, $2$ or $4$ dependent on whether $e_1=1$, $2$, or $\ge3$ respectively, we have four cases to consider:
Case 1: If $e_1=0$ there are only odd primes $p_i$ each factoring in $N_i=2$ classes in the product $$N=N_1\cdot N_2\dotsm N_k=2\cdot 2\dotsm 2=2^{k}\tag{1}$$
Case 2: If $e_1=1$ there are $$N=N_1\cdot N_2\dotsm N_k=1\cdot N_2\dotsm N_k=1\cdot2^{k-1}=2^{k-1}\tag{2}$$ many classes of solution.
Case 3: If $e_1=2$ there are $$N=N_1\cdot N_2\dotsm N_k=2\cdot N_2\dotsm N_k=2\cdot2^{k-1}=2^k \tag{3}$$ many classes of solution.
Case 4: If $e_1\geq3$ there are $$N=N_1\cdot N_2\dotsm N_k=2^2\cdot N_2\dotsm N_k=2^2\cdot2^{k-1}=2^{k+1} \tag{4}$$ many classes of solution.
Now see how these results generalise to $n$.
If $n$ is odd then we are in case $(1)$.
If $n\equiv2\pmod{4}$ then $n$ has $2$ as a prime factor of power $1$, i.e., $e_1=1$, then $N=2^{k-1}$, and we are in case $(2)$.
If $n$ is odd or $e_1=2$ then $N=2^k$, and we are in case $(3)$.
If $n$ is divisible by $8t$, for some $t\in\mathbb{Z}$, then $N=2^{k+1}$ and we are in case $(4)$.