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I'm trying to prove this statement. I've already checked here in StackExchange but I can only find the proof of the converse.

In Wikipedia, it is stated: "A number $a$ relatively prime to an odd prime $p$ is a residue modulo any power of $p$ if and only if it is a residue modulo $p$", and I want to prove the only if part.

This problem appeared to me when I wanted to solve the following problem:

Check that the only solutions, up to congruence, of $X^2 \equiv 25 $ (mod $p^k)$, for any $k$ and for any prime $p \neq 2$, $5$, are $X \equiv \pm 5 $ (mod $p^k)$.

Any help would be appreciated.

Bill Dubuque
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Albert
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    This result (its generallization, to be precise) is called Hensel's lemma. See here https://en.m.wikipedia.org/wiki/Hensel%27s_lemma – richrow Feb 15 '21 at 19:07
  • Thank you for your comment, but at the time this exercise was proposed to me, I hadn't already seen that Lemma, it only comes afterwards, so I shall not use it here, there has to be another way. – Albert Feb 15 '21 at 19:15
  • The title and the body say you want to prove different implications. I assume the body is correct. – rogerl Feb 15 '21 at 19:17
  • Thank you for noticing that, I already corrected the body, clearly it was not correct due to my previous explanation. – Albert Feb 15 '21 at 19:19
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    OK, you can prove it in this particular case. Suppose we know that there is solution $x\equiv a\pmod {p^k}$ and we want to build solution modulo $p^{k+1}$. Consider $x_0\equiv a+tp^k$ and try to find $t$ such that $x_0$ is a solution modulo $p^{k+1}$. – richrow Feb 15 '21 at 19:21
  • I appreciate your help, but I corrected a few moments earlier that that is not the implication that I want to prove. – Albert Feb 15 '21 at 19:22
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    If $a \equiv b^2 \pmod{p^k}$, you can just reduce both sides mod $p$ and get $\overline{a} \equiv \overline{b}^2 \pmod{p}$, where $\overline{a}$ and $\overline{b}$ are the reductions mod $p$. More formally, there is the reduction homomorphism $\pi: \mathbb{Z}/p^k\mathbb{Z} \to \mathbb{Z}/p\mathbb{Z}$ taking an element to its reduction mod $p$. If $a = b^2$ in $\mathbb{Z}/p^k\mathbb{Z}$, then $\pi(a) = \pi(b^2) = \pi(b)^2$ in $\mathbb{Z}/p\mathbb{Z}$. – Viktor Vaughn Feb 15 '21 at 19:32
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    If you are trying to prove the direction indicated in the question, note that if $x\equiv a^2\mod{p^k}$, then $x-a^2 = tp^k = tp^{k-1}p$, so that $x\equiv a^2\mod{p}$. – rogerl Feb 15 '21 at 19:35
  • These last two comments do the job, thank you all ! – Albert Feb 15 '21 at 19:41
  • @Albert This is a special case of basic divisibility theorems in $\Bbb Z$ (true in any UFD), as I explain in my answer. – Bill Dubuque Feb 15 '21 at 22:24

2 Answers2

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You indicated that the problem

    Let $k$ be a positive integer and let $p$ be a prime other than $2$ or $5$.

    Show that the only solutions, up to congruence, of $x^2 \equiv 25\;(\text{mod}\;p^k)$ are $x \equiv \pm 5\;(\text{mod}\;p^k)$.

is the underlying problem which you are trying to solve.

For the above problem, there is no need to consider quadratic residues.

Instead, we can argue as follows . . .

Suppose $x$ is an integer such that $x^2 \equiv 25\;(\text{mod}\;p^k)$.

Note that $x+5$ and $x-5$ can't both be divisible by $p$, else their difference $$ (x+5)-(x-5)=10 $$ would be divisible by $p$, contrary to $p\ne 2,5$. \begin{align*} \text{Then}\;\;& x^2 \equiv 25\;(\text{mod}\;p^k) \\[4pt] \implies\;& p^k{\,\mid\,}x^2-25 \\[4pt] \implies\;& p^k{\,\mid\,}(x+5)(x-5) \\[4pt] \implies\;& p{\,\mid\,}(x+5)(x-5) \\[4pt] \implies\;& p{\,\mid\,}(x+5)\;\text{or}\;p{\,\mid\,}(x-5)\;\text{but not both} \\[4pt] \end{align*} Then since $p^k{\,\mid\,}(x+5)(x-5)$ and exactly one of $x+5,x-5$ is divisible by $p$, it follows that exactly one of $x+5,x-5$ is divisible by $p^k$.

Therefore $x \equiv \pm 5\;(\text{mod}\;p^k)$, as was to be shown.

quasi
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  • @Albert The above proof is a special case of basic divisibility theorems in $\Bbb Z$ (true in any UFD), as I explain in my answer. – Bill Dubuque Feb 15 '21 at 22:24
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This is a special case of the primary (prime power) form of Euclid's Lemma below (put $\,B=x\!+\!5,\,C=x\!-\!5,\,$ so $\,(p,B,C) = (p,B,B\!-\!C) = (p,x\!+\!5,10)=1\,$ by $\,(p,10)=1)$

Primary Euclid Lemma $\ \color{#0a0}{(p,B,C)=1},\ p^k\mid BC\Rightarrow p^k\mid B\,$ or $\,p^k\mid C,\ $ for prime $p$

which is the special case $\,a\,$ is primary in a primitive primal form of Euclid's Lemma

$\rm\color{#0a0}{Primitive}$ Primal Euclid's Lemma $\ $ If $\,a,B,C\,$ are integers then

$$\begin{align} &a\mid BC,\ \color{#0a0}{(a,B,C)=1}\\[.2em] \Longrightarrow \ &a = bc,\qquad\:\! \color{#0a0}{(b,c)=1}\\ &b\mid B,\,c\mid C \end{align}$$

Proof $\,\ a = bc,\ b\mid B,\ c\mid C\,$ follows by the link (by gcd laws or $\Bbb Z$ a UFD) and the coprimality is clear: $\ d\mid b,c\,\Rightarrow d\mid B,C,bc\!=\!a\,$ so $\,d\mid\color{#0a0}{(B,C,a)=1}$

Bill Dubuque
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