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I know that $\{a_i\}=R(pq)$, and the title is step $(b)$, here is step $(a)$ (maybe a hint?):

Let $p$ and $q$ be two distinct odd primes.

$(a)$ Show that all the solutions of the congruence $x^2 \equiv 1 \pmod {pq}$ are given by
$x \equiv \{\;1,\;-1,\;p^{q-1}-q^{p-1},\;q^{p-1}-p^{q-1}\}\pmod{pq}$.

I can prove $(a)$, but cannot prove $(b)$ which is a generalization of Wilson's theorem.

Bill Dubuque
  • 272,048

2 Answers2

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As explained here, by pairing up inverses the product reduces to the product of all self-inverse $a_i$ (roots of $\,x^2\equiv 1).\, $ By CRT the roots are your $\,x\equiv (1,1),(-1,-1),\color{#c00}{(-1,1)},\color{#0a0}{(1,-1)}\pmod{p,q}\,$ with product $(1,1),\,$ which maps to $1\!\pmod{\!pq}.\ $ QED

Remark $ $ As explained in the link, the same proof generalizes Wilson's theorem to $\,\Bbb Z_n\,$ for odd $n$ having at at least two distinct prime factors. I explain in that answer how it generalizes even further, e.g. if a finite abelian group has a unique element of order $2$ then it is the product of all the elements; otherwise the product is $1$. There are motley twists on results like this - some well-known - some not. Follow said link to learn more.

Bill Dubuque
  • 272,048
  • Note $ \bmod (p,q)!:,\ u := p^{q-1}!-q^{p-1}\equiv\color{#c00}{(-1,1)},$ and $ , q^{p-1}!-p^{q-1}\equiv -u\equiv \color{#0a0}{(1,-1)}\ \ $ – Bill Dubuque Mar 27 '20 at 17:49
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Claim: $\{ a_i \} \pmod{p}$ (as a multi-set) consists of $ (q-1)$ 1's, $ (q-1)$ 2's, $ (q-1)$ 3's, $\ldots$, $ (q-1)$ $p-1$'s.

Proof: Let's count how many times 1 appears.
Consider $ kp + 1 $ where $ k = 0 $ to $q-1$.
Exactly 1 of them is a multiple of $q$, which isn't in $R(pq)$.
Hence, there are $(q-1) 1's$ in $ \{ a_i \} \pmod{p}$.
Similarly for the other residues.

Claim: $ \prod a_i \equiv 1 \pmod{p}$.

Proof: From the above, $ \prod a_i \equiv (\prod_{i=1}^{p-1} i )^{q-1} \equiv (-1)^{q-1} \equiv 1 \pmod {p}$
Here we use 1) $p$ is prime and we apply Wilson's theorem, 2) $q$ is odd so $q-1$ is even.

Corollary: $ \prod a_i \equiv 1 \pmod{pq}$.

With reference to Bill's comment, this also generalizes to showing that for $p, q,r$, odd distinct primes, and $\{a_i\}$ the residue class of $pqr$,

$$\prod a_i \equiv 1 \pmod{pqr}$$

Calvin Lin
  • 68,864
  • This ignores the hint given in the OP's textbook, and likely doesn't generalize (but we can't be sure since so little is said we have no idea precisely what type of proof is intended) – Bill Dubuque Mar 27 '20 at 16:29
  • It's just part (a), which need not necessarily be a hint. I'm not sure what you mean by "likely doesn't generalize". My interpretation of OP's statement is that "b) generalizes Wilson's theorem". – Calvin Lin Mar 27 '20 at 16:45
  • As a side note, this approach does generalize to $ \prod a_i \equiv 1 \pmod{pqr}$ for odd primes. – Calvin Lin Mar 27 '20 at 16:47
  • I can't read your mind. Precisely what is the proof you intend and how does it generalize? Please be more precise (there are various ways to prove it). – Bill Dubuque Mar 27 '20 at 16:53
  • I've been repeatedly told not to post complete answers on MSE when OP shows no work. So, I'm not sure what to do right now. – Calvin Lin Mar 27 '20 at 16:55
  • We've had many problems in the past with users posting "hints" that lead nowhere. You have to give enough details to ensure that we know what you intend. If someone asks you to elaborate then you should do so. Otherwise there is no way to distinguish math from pseudo-math. – Bill Dubuque Mar 27 '20 at 17:00
  • I can promise that my hints in solutions are proper hints. I've written it up completely now. – Calvin Lin Mar 27 '20 at 17:01
  • Ok, now it is clearer what you intend in the OP's case. But is not clear precisely how you intend to generalize that from $pq$ to arbitrary $n$, or to arbitrary finite abelian groups (the method hinted to the OP - though not as clear as it could be - does in fact generalize that widely, e.g. follow the link in my answer). – Bill Dubuque Mar 27 '20 at 17:39
  • And I don't follow why you think the above uses the hint. Where do you use part (a) in your argument? – Bill Dubuque Mar 27 '20 at 17:41
  • I am not generalizing to "arbitrary $n$". E.g. It is not true for $n = p^2$. It can be generalized to arbitrary product of distinct primes, through the exact same approach. 2) There is no requirement to use part (a) in the argument. OP says "maybe a hint" and not "please use this". (Or at least, that's my interpretation esp for a new user who posts with title as a problem statement, then the question is additional details)
    • – Calvin Lin Mar 27 '20 at 20:13
  • Maybe the idea you have in mind does not generalize (again I can't be sure since you give no details). But the part $(a)$ hinted method does generalize to all $n$ and also to finite abelian groups - as I mentioned above. It seems clear to me that part $(a)$ was meant to be used in a proof like that I mention in my answer. What would be the point of it otherwise? – Bill Dubuque Mar 27 '20 at 20:49
  • To be precise, the group theoretic generalization I refer to is: if a finite abelian group has a unique element of order $2$ then it equals the product of all the elements; otherwise the product is $1$. There are various twists on results like this - some well-known - some not. Follow the link in my answer to learn more. – Bill Dubuque Mar 27 '20 at 21:13
  • Thanks a lot. Maybe I can explain the first claim by: Consider a map from R(mn) to R(m) *R(n) (Cartesian product) by from a to (the remainder of a divided by q,the remainder of a divided by q) which is one-one and onto. Then we can get the conclusion of the first claim: {a_i} (mod p) =1,2,3...p-1. And the total # of {a_i}=(p-q)(q-1). –  Mar 29 '20 at 05:36