I was reading the documentation of a least square function of numpy library at this link, I found written
Solves the equation $a x = b$ by computing a vector $x$ that minimizes the squared Euclidean $2$-norm
But I've never heard about squared Euclidean $2$-norm, I know there are squared $L^2$-norm vectors, is it also a valid terminology to say only $2$-norm?
The phrase is, to my ear, redundant, but not wrong. In particular, there is an important family of (topologically equivalent) norms which are defined on $\mathbb{R}^n$. For any $p \in [0,\infty)$ and for any $x = (x_1, \dotsc, x_n) \in \mathbb{R}^n$, define $$ \|x\|_{p} := \left( |x_1|^p + \dotsb + |x_n|^p \right)^{1/p}. $$ The quantity $\|x\|_p$ is called the $p$-norm, or the $L^p$-norm, of $x$. Taking $p=2$ in this formula gives $$ \|x\|_2 = \left( |x_1|^2 + \dotsb + |x_n|^2 \right)^{1/2}, $$ which is the $2$-norm (or $L^2$-norm) of $x$.
On the other hand, the ancients had a technique for computing the distance between two points in $\mathbb{R}^n$ which amounts to a generalized Pythagorean theorem. Specifically, if $x,y \in \mathbb{R}^n$, then the distance between $x$ and $y$ is given by $$ d(x,y) = \sqrt{ (x_1 - y_1)^2 + \dotsb + (x_n - y_n)^2 }. $$ This distance makes sense in the context of Euclidean geometry, thus this is often referred to as the Euclidean distance. Moreover, every distance induces a norm, so from this formula we get the Euclidean norm, which is defined by $$ \|x\|_{\text{Euclid}} := d(x,0) = \sqrt{(x_1-0)^2 + \dotsb + (x_n-0)^2} = \sqrt{x_1^2 + \dotsb + x_n^2} $$ for each $x\in \mathbb{R}^n$.
Notice that the Euclidean norm is precisely the $2$-norm. Thus it is not incorrect to say "the Euclidean $2$-norm". Again, I find such language to be redundant, but this redundancy does, perhaps, serve to clarify some.
Finally, in many applications, the goal is to minimize the distance between two points, or to minimize the length of some vector. Indeed, this is precisely what the function described in the question seeks to do: it solves the system $ax = b$ by minimizing the length of $b-ax$ with respect to the Euclidean norm. However, note that $$ \text{$\|b-ax\|_{2}$ is minimal} \iff \text{$\|b-ax\|_{2}^{2}$ is minimal}. $$ Thus the function actually minimizes the square of the Euclidean $2$-norm. Or, in slightly fewer letters, the function minimizes the squared Euclidean $2$-norm.