12. Proof: The sufficiency has been proved in the previous problem. Below, we only prove the necessity.

Let $m \geqslant 3, (a, m)=1$, and $x^{2} \equiv a(\bmod m)$ has exactly two solutions. We need to prove that $\boldsymbol{m}$ must have a primitive root. We use proof by contradiction. Assume $m$ does not have a primitive root. From the discussion in Chapter $\S 1$ —§4, $m$ must not be of the following forms:
$$2,4, p^{s}, 2 p^{s} \quad(p \geqslant 3, s \geqslant 1)$$

Thus, $m \geqslant 5$, and $m$ must be one of the following forms:
(1) $m=2^{b}, b \geqslant 3$,
(2) $m=2^{c} p^{x}, \alpha \geqslant 1, c \geqslant 2$,
(3) $m=2^{c} p_{1}^{x_{1}} \ldots p_{s}^{x_{s}}, c \geqslant 0, s \geqslant 2, \alpha_{\mathrm{i}} \geqslant 1(1 \leqslant i \leqslant s), p_{i} \geqslant 3(1 \leqslant i \leqslant s)$.

First, consider case (1). In this case,
$$x^{2} \equiv 1\left(\bmod 2^{b}\right)$$

Besides $x \equiv 1, -1$, there are at least two other solutions $x \equiv 2^{b-1}-1, 2^{b-1}+1$, which are clearly pairwise incongruent $\left(\bmod 2^{b}\right)$. Therefore, case (1) is impossible.

Next, consider case (2). Since $c \geqslant 2$, we have
$$x^{2} \equiv 1(\bmod 2)$$

which has at least two solutions $x_{1}, x_{2}$ that are incongruent $\bmod 2$.
We also know that
$$x^{2} \equiv 1\left(\bmod p^{x}\right)$$

has two solutions $y_{1}, y_{2}$ that are incongruent $\bmod p^{\alpha}$. Solving
(using the Chinese Remainder Theorem) gives
$$x^{2} \equiv 1\left(\bmod 2^{c} p^{x}\right)$$

four solutions $\left(\bmod 2^{c} p^{x}\right)$, which are clearly pairwise incongruent $\left(\bmod 2^{c} p^{x}\right)$. Therefore, case (2) is also impossible.

Finally, consider case (3). Using the same method as in case (2), we can prove that in this case,
$$x^{2} \equiv 1(\bmod m)$$

has more than 2 solutions that are pairwise incongruent $(\bmod m)$. Therefore, case (3) is also impossible. In any other case, $m$ must have a primitive root, which completes the proof.
$$\begin{array}{l} \left\{\begin{array} { l } { \overline { x } _ { 3 } \equiv x _ { 2 } ( \operatorname { m o d } 2 ^ { c } ) , } \\ { \overline { x } _ { 3 } \equiv y _ { 1 } ( \operatorname { m o d } p ^ { x } ) , } \end{array} \quad \left\{\begin{array}{l} \bar{x}_{4} \equiv x_{2}\left(\bmod 2^{c}\right), \\ \bar{x}_{4} \equiv y_{2}\left(\bmod p^{x}\right), \end{array}\right.\right. \end{array}$$