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I've been dividing up the elements of reduced residue system relative a prime $p_n$ into congruence classes modulo $p_{n+1}$ and I noticed that each congruence class is represented.

If $r$ = the number of elements in a reduced residue system relative $p_n\#$, I am finding that for each congruence class modulo $p_{n+1}$, there are at least $1+\left\lfloor\dfrac{r}{3p_{n+1}}\right\rfloor$ instances of each congruence class.

For $5\#$, there are $8$ elements and we see at least $1$ of each relative $7$: $(0,\{7\}), (1,\{1,29\}), (2,\{23\}), (3,\{17\}), (4,\{11\}), (5,\{19\}), (6,\{13\})$

For $7\#$, there are $48$ elements and we see at least $3 > 1+\left\lfloor\frac{48}{33}\right\rfloor$ for each relative $11$: $(0,\{11,121,143\}), (1,\{1,23,67,89,169,197\}), (2,\{13,79,101,113,191\}), (3,\{47,157,199\}), (4,\{37,59,103,179,193\}), (5,\{137,173,187\}), (6,\{17,61,83,127,149,181,209\}), (7,\{29,73,139\}), (8,\{19,41,107,151\}), (9,\{31,53,97,163\}), (10,\{43,109,131\})$

I wrote a computer application and confirmed this up to $23\#$.

Is this always true?

Edit 1: Changed the term "relative" to "modulo" for congruence classes to make the question more clear.

Edit 2: I wrote a simple application and identified the following for primes in relation to the reduce residue system relative $7\#$

Let $R_{7\#}$ be the reduced residue system relative $7\#$

  • $R_{7\#}$ contains $2$ sets of each distinct congruence classes modulo $13$
  • $R_{7\#}$ contains $2$ sets of each distinct congruence classes modulo $17$
  • $R_{7\#}$ contains $2$ sets of each distinct congruence classes modulo $19$
  • $R_{7\#}$ contains $1$ set of each distinct congruence class modulo $23$
  • $R_{7\#}$ contains $1$ set of each distinct congruence class modulo $29$
  • $R_{7\#}$ contains $1$ set of each distinct congruence class modulo $31$
  • For all other primes $> 31$, $R_{7\#}$ does not contain each distinct congruence classes. For primes $> 209$, $R_{7\#}$ maps to $48$ distinct congruence classes only.

Edit 2: I believe that I am starting to make progress in thinking about this question. Here are some thoughts. Please let me know if any of these ideas are wrong or unclear.

  • Let $R_{p_i\#}$ be the reduced residue system relative $p_i\#$.
  • Let $R_{p_{i-1}\#}$ be the reduced residue system relative $p_{i-1}\#$

  • Let $p_k$ be any prime where $k > i$

  • $1, 1+p_{i-1}\#, 1+2p_{i-1}\#, \dots, 1+(p_k-1)p_{i-1}\#$ forms a complete residue system modulo $p_k$ and can be continued as long as we want to form a sequence of $n$ complete residue systems modulo $p_k$

  • Label these elements $c_1 = 1, c_2 = 1+p_{i-1}\#, c_3, \dots , c_{p_k}=1+(p_k-1)p_{i-1}\#, \dots, c_{p_k+1}=1+p_k p_{i-1}\#, \dots, c_{n p_k}=1+(n p_k-1)p_{i-1}\#$
  • Using Euler's totient function $\varphi(x)$, there are $\varphi(p_{i-1}\#)$ elements in $R_{p_{i-1}\#}$

  • The elements of $R_{p_{i-1}\#}$ can be mapped to an element $c_i$ such that $c_{r_1}, c_{r_2}, \dots, c_{r_\varphi(p_{i-1}\#)}$ where $r_1 < r_2 < r_3 < \dots < r_{\varphi(p_{i-1}\#)} < r_{\varphi(p_{i-1}\#)+1}$ where $c_{r_1} \equiv c_{r_{\varphi(p_{i-1}\#)+1}} \pmod p_k$

  • If for any $r_i$, there is $r_{i+1} - r_i > p_i+1$, then $R_{p_i\#}$ does not contain every distinct congruence class modulo $p_k$

Larry Freeman
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    What do you mean "relative $p_{n+1}$"? – Konstantinos Gaitanas Mar 11 '15 at 12:05
  • I meant that the elements of the reduced residue system relative $p_{n}$ can be divided into congruence classes modulo $p_{n+1}$. I have changed the wording. Thanks for noticing! – Larry Freeman Mar 11 '15 at 14:00
  • Larry, the number of edits you made triggered a system flag. This caused no harm in your case, but I would like to advice you of the fact that sometimes multiple edits annoy other users. See this recent meta thread as well as older linked meta threads for a discussion as to why that is so, and what can you do to the alleviate the problem. If you are now happy with the formulation of the question, then this is no concern to you whatsoever. Just FYI. – Jyrki Lahtonen Mar 14 '15 at 06:40
  • Hi Jyrki, thanks for letting me know. I was trying to make progress on the question myself. I will not make any more edits. – Larry Freeman Mar 14 '15 at 08:13

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