12. In the sequence $\left\{a_{n}\right\}$, if $a_{n}^{2}-a_{n-1}^{2}=p\left(n \geq 2, n \in \mathbf{N}^{*}, p\right.$ is a constant), then $\left\{a_{n}\right\}$ is called a "sequence of equal variances". The following are judgments on the "sequence 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 $\left\{a_{k n}\right\}\left(k \in \mathbf{N}^{*}, k\right.$ 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 the correct propositions, the sequence numbers are $\qquad$ (fill in the sequence numbers of all correct propositions on the line)
Rong Case 1234.
Analysis 1 Because $\left[(-1)^{n}\right]^{2}-\left[(-1)^{n-1}\right]^{2}=0$, the sequence $\left\{(-1)^{n}\right\}$ meets the definition of an "equal variance sequence"; 2 According to the definition, it is obvious that $\left\{a_{n}^{2}\right\}$ is an arithmetic sequence;
$3 a_{k n}^{2}-a_{k(n-1)}^{2}=a_{k n}^{2}-a_{k n-1}^{2}+a_{k n-1}^{2}-a_{k n-2}^{2}+\cdots+a_{k n-k+1}^{2}-a_{k(n-1)}^{2}=k p$ meets the definition;
4 If the sequence $\left\{a_{n}\right\}$ satisfies $a_{n}^{2}-a_{n-1}^{2}=p, a_{n}-a_{n-1}=d$ (where $d$ is a constant):
If $d=0$, it is clear that $\left\{a_{n}\right\}$ is a constant sequence;
If $d \neq 0$, then dividing the two equations yields $a_{n}+a_{n-1}=\frac{p}{d}$, so $a_{n}=\frac{d^{2}+p}{2 d}$ (a constant), meaning $\left\{a_{n}\right\}$ is a constant sequence.