12. In the sequence $\left\{a_{n}\right\}$, if $a_{n}^{2}-a_{n-1}^{2}=p(n \geqslant 2, n \in \mathbf{N}, p$ is a constant $)$, then $\left\{a_{n}\right\}$ is called a "sequence of equal variances". The following are judgments on sequences of equal variances:
(1) The sequence $\left\{(-1)^{n}\right\}$ is a sequence of equal variances;
(2) If $\left\{a_{n}\right\}$ is a sequence of equal variances, then $\left\{a_{n}^{2}\right\}$ is an arithmetic sequence;
(3) If $\left\{a_{n}\right\}$ is a sequence of equal variances, then $\left\{a_{k n}\right\}(k \in$ $\mathbf{Z}_{+}, k$ is a constant $)$ is also a sequence of equal variances;
(4) If $\left\{a_{n}\right\}$ is both a sequence of equal variances and an arithmetic sequence, then the sequence is a constant sequence.
Among these, the correct proposition numbers are . $\qquad$
12. (1)(2)(3)(4).
(1) Since $\left((-1)^{n}\right)^{2}-\left((-1)^{n-1}\right)^{2}=0$, therefore, $\left\{(-1)^{n}\right\}$ satisfies the definition.
(2) According to the definition, it is obvious that $\left\{a_{n}^{2}\right\}$ is an arithmetic sequence.
$$
\text { (3) } a_{k n}^{2}-a_{k(n-1)}^{2}=\sum_{i=1}^{k}\left(a_{k n+1-i}^{2}-a_{k n-i}^{2}\right)=k p,
$$

satisfies the definition.
(4) If the sequence $\left\{a_{n}\right\}$ satisfies:
$$
a_{n}^{2}-a_{n-1}^{2}=p, a_{n}-a_{n-1}=d \text { ( } d \text { is a constant), }
$$

then when $d=0$, it is obvious that $\left\{a_{n}\right\}$ is a constant sequence;
when $d \neq 0$, dividing the two equations yields
$a_{n}+a_{n-1}=\frac{p}{d} \Rightarrow a_{n}=\frac{d^{2}+p}{2 d}$ (a constant),

i.e., $\left\{a_{n}\right\}$ is a constant sequence.