# Existence of a non-square integer $L$ such that $L$ is a quadratic residue modulo $p^n$ for all $n$

The background of this question is on the sequences in $p$-adic integers, in which I originally looked into the following problem:

Can a sequence of perfect squares converge to a non-square value under the $p$-adic metric?

Where the $p$-adic metric is, as usual, defined as $$d_p(m,n)=p^{-\max\{n\in\mathbb N\colon p^n|(m-n))\}}$$ For $m\neq n$, and $d_p(m,m):=0$

Obviously, it is ask for the existence of a function $f:\mathbb N\to\mathbb Z$ and a non-square number $L$ such that for all $n\in\mathbb N$, $$f(n)^2-L=p^{g(n)}q(n)$$ Where $g:\mathbb N\to\mathbb N$ is an unbounded function and $q(n)$ is coprime to $n$ for all $n$. Clearly, this is also equivalent to say that the equation $$x^2\equiv L\pmod{p^n}$$ Is soluble for infinitely many $n$. But we know that if $L$ is a quadratic residue modulo $p^n$, it is also a quadratic residue modulo $p^{n-1}$. Therefore, the statement resolves to the following proposition:

There is (not) a non-square integer $L$ such that it is a square modulo $p^n$ for every $n\in \mathbb N$.

Which I know neither is true or not, nor how to start off decently. Though I have not found any of them, and I believe that such $L$ does not exist, i still cannot find out a complete proof. The proposition can indeed be generalized into the following

There is (not) a non-$k$th power $L$ such that it is congruent to a $k$th power modulo $p^n$ for all $n\in\mathbb N$

Could you give me a hint to start off? Thanks in advance.

Not only are there plenty of non-squares which become squares in the $p$-adic numbers $\mathbf{Z}_p$, the phenomena underlying this was one of the motivating reasons for constucting the $p$-adic numbers in the first place. Here are some hints (only hints!) to get you started.

Notice that

$$2^2 = -1 \mod 5,$$ $$(2 + 5)^2 = 7^2 = -1 \mod 5^2,$$ $$(2 + 5 + 2 \cdot 5^2)^2 = 57^2 = -1 \mod 5^3.$$

So now try the following:

A. Modify $57$ by adding a multiple of $5^3 = 125$ so that $(57 + k \cdot 5^3)^2 \equiv -1 \mod 5^4$

B. Inductively construct $x_n$ with the following properties.

1. $x_1 = 2$,
2. $x_n \equiv x_{n-1} \mod 5^{n-1}$,
3. $x^2_n \equiv -1 \mod 5^{n}$.

C. Think about how general the construction above is. What would happen if one replaces $-1$ and $5$ by $a$ and $p$, where $a$ is a square modulo $p$, and $p$ is an odd prime. What happens when $p = 2$?