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 $p = 2$?

| cite | improve this answer | |

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.