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Let $\lambda$ be a given integer and $p$ is a prime and integer $r$.

In this case, for $\lambda$ why are there $p^{r-1}$ number of possible $\mu$ from the residue class $p^r$ that satisfies $\mu \equiv \lambda$ mod $p$ and $p^{r-2}$ number of possible $\mu$ that satisfies $\mu \equiv \lambda$ mod $p^2$?

Now I want to show this because I want the result that for any given integers $(\lambda_1, \dots, \lambda_r)$, there are $\Pi_{j=1}^{r-1}p^{r-j}=p^{r(r-1)/2}$ different vectors $(\mu_1,\dots, \mu_r)$ modulo $p^r$ such that $$\mu_j \equiv \lambda_j\; \text{mod} \;p^j \; \forall 1\le j \le r.$$

Edit This question was written incorrectly so I've edited it and lulu's comments have solved the reasoning.

Bill Dubuque
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nomadicmathematician
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  • Not following. Why does it matter if $p>r$? And $\mu \equiv \lambda \pmod {p^r}\implies \mu \equiv \lambda \pmod {p}$ so why list the redundant condition? – lulu Jul 24 '23 at 13:47
  • I suggest: provide an explicit example of what exactly it is you are trying to count. – lulu Jul 24 '23 at 13:49
  • @lulu $p>r$ is irrelevant for this it was just a part of assumption of the Lemma I am trying to prove. I've provided the context in which I need this result. If this is still unclear I will state the whole lemma. – nomadicmathematician Jul 24 '23 at 13:55
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    So, delete the irrelevant assumption. And...why impose redundant conditions? If $p^r$ divides $n$ then $p^j$ divides $n$ for all $j<r$, so what are you saying? Again, I suggest providing an explicit numerical example of whatever it is you are trying to describe. – lulu Jul 24 '23 at 14:00
  • To stress: what you have written is very unclear. It's not even clear what $\mu$ is supposed to be. If $\mu$ can be any integer then, presumably, there are infinitely many $\mu$ congruent to a given integer mod whatever you like. If $\mu$ is meant to be a residue $\pmod {p^r}$ then there is only one choice of $\mu$ congruent to a given integer $\pmod {p^r}$. But maybe $\mu$ is something else altogether? As I say, an explicit example illustrating the count you have in mind should clarify your meaning. – lulu Jul 24 '23 at 14:10
  • Voting to close the question as it is not clear what you are asking. If you can, please edit your post for clarity. Examples are always a good idea. – lulu Jul 24 '23 at 14:22
  • @lulu Okay thank you for the clarification. This was all a misunderstanding of the statement from me. So I believe what is meant here is that in the residue class $p^r$ there are at most $p^{r-1}$ numbers $\mu$ that is congruent to a given $\lambda$ mod $p$ and $p^{r-2}$ numbers of $\mu$ that is congruent to a given $\lambda$ mod $p^2$ and so on. Does this make sense? – nomadicmathematician Jul 24 '23 at 14:27
  • @lulu so for example if $r=2, p=3$ and $\lambda_1 = 7 , \lambda_2=12$ then there are three $\mu_1=1,4,7$ and one $\mu_2=3$ choices that makes sense. – nomadicmathematician Jul 24 '23 at 14:31
  • $p=4$ is not a prime. And, as a general rule, it's best to avoid $2$ as a choice of a "random" prime. Lots of things that are true for odd primes work poorly for $2$. – lulu Jul 24 '23 at 14:32
  • @lulu That was a typo from me sorry. Anyways your pointing out of my mistakes made everything clear. I will just close this question sorry for the inconvenience. – nomadicmathematician Jul 24 '23 at 14:35
  • All that said: There are $p^{r-1}$ multiples of $p$ $\pmod {p^r}$. They are ${0, 1\times p, 2\times p, \cdots,( p^{r-1}-1)\times p}$. I think this is what you are asking, no? – lulu Jul 24 '23 at 14:36
  • @lulu I think that is right. So if $\lambda_1$ is a given integer then it will be congruent to some number between 0 and p-1. And we have $p^{r-1}$ multiples which will be in the residue class mod $p^r$ and those are as you stated in the parentheses. – nomadicmathematician Jul 24 '23 at 14:59
  • Yes, that's my point. I think that's what you have in mind here, no? – lulu Jul 24 '23 at 15:00
  • I don't follow your question, but with regard to the query in the title: Keeping to the convention that a number modulo $q$ is smaller than $q$ (are you using that understanding in your question?), numbers of the form $kp^r+j$ have residues $\bmod p^r$ that are larger than $p$ when $j>p$. The only numbers that have equal residues both $\bmod p$ and $\bmod p^r$ have the form $kp^r+j$ for $j\le p$. Of course, there are infinitely many of these because $k$ can have any integer value. – Keith Backman Jul 24 '23 at 15:01
  • @KeithBackman Yes the question was incorrect and the discussion with lulu has solved it. – nomadicmathematician Jul 24 '23 at 15:12
  • The reduction map from $!\bmod \color{#c00}kn,$ to $!\bmod n,$ is $,\color{#c00}k$-to-$1.,$ Here $,nk = p^r,\ n = p^2$ so $,\color{#c00}k = p^r/p^2 = p^{r-2}\ \ $ – Bill Dubuque Jul 24 '23 at 15:18

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