Recurring decimals in prime fractions

[Oxymoron]

And with per(1/p^2) we can say that it too is a divisor of phi(p^2). Which, as you have already said, means that per(1/p^2) divides p*(p-1)

 

[Oxymoron]

Posted by Shmoe

Yes, but you can also say something about per(1/pq) in terms of per(1/p) and per(1/q), namely per(1/pq) = lcm(per(1/p),per(1/q))

I thought we came to the conclusion that this wasnt true. It was because per(1/7) happened to equal phi(7), and per(1/17) happened to equal phi(17). Now you are saying that it was correct.

 

[shmoe]

I thought we came to the conclusion that this wasnt true. It was because per(1/7) happened to equal phi(7), and per(1/17) happened to equal phi(17). Now you are saying that it was correct.

lcm(phi(p),phi(q)) and lcm(per(1/p),per(1/q)) are not always the same. Look back to posts 25/26.

 

[Oxymoron]

SUMMARY:

1. The period of 1/pq equals the order of 10 mod pq.
2. Let t=ord_(pq)(10).
3. By definition of order, we must have 10^t=1(mod pq).
4. This implies two linear congruences
a) 10^t = 1(mod p)
b) 10^t = 1(mod q)
5. We know that t divides ord_p(10) <=> 10^t=1(mod p)
6. We know that t divides ord_q(10) <=> 10^t=1(mod q)
7. We also know that ord_p(10) divides p-1 (or in other words phi(p)).
8. From 4a and 4b we know that if t divides two different numbers then it must divide their lowest common multiple. That is, t must divide lcm(ord_p(10),ord_q(10)).
9. Which means that t divides lcm(phi(p),phi(q)).

How does this look?

 

[Oxymoron]

It doesn't look right at all. t divides ord_p(10) does not mean that t divides phi(p). ...(or does it?)

 

[shmoe]

5. We know that t divides ord_p(10) <=> 10^t=1(mod p)
6. We know that t divides ord_q(10) <=> 10^t=1(mod q)

These are backwards.

ord_p(10) divides t<=> 10^t=1(mod p)

7. We also know that ord_p(10) divides p-1 (or in other words phi(p)).

This is true, but not necessary. You have a phi fixation.

8. From 4a and 4b we know that if t divides two different numbers then it must divide their lowest common multiple. That is, t must divide lcm(ord_p(10),ord_q(10)).

This is backwards. You'll have ord_p(10) and ord_q(10) dividing t, and hence so does their least common multiple.

In the direction you are trying to go, you need something like (4)=>(3), which follows when p and q are distinct primes (hence relatively prime). This will let you show the order of 10 mod pq divides the order of of 10 mod p and 10 mod q, so it divides their least common multiple.

9. Which means that t divides lcm(phi(p),phi(q)).

Again true, but saying t divides lcm(per(1/p),per(1/q)) is in general stronger.

 

[Oxymoron]

Thankyou Shmoe for all your help and patience. I believe I can now write a correct solution.

 

[bdeln]

**bump**

I'm working on a similar problem, ie, I need to find the period of 1/pq if (p,q) = (p,10) = (q,10) = 1.

If the period of 1/p is r and the period of 1/q is s, then

$10^r \equiv 1 (p) \mbox{ and } 10^s \equiv 1 (q)$.

So, let t be the period of 1/pq, ie, $10^t \equiv 1 (pq)$, then $10^t \equiv 1 (p) \mbox{ and } 10^t \equiv 1 (q)$.

Now $r|t$ and $s|t$ (since if the order of a mod n is k and $a^t \equiv 1 (n)$, then k|t). This implies that $\textstyle \left. \frac{rs}{\mbox{gcd}(r,s)} \right| t \Rightarrow \mbox{lcm}(r,s)|t.$

I just have to show that $t|\mbox{lcm}(r,s)$ so I can say the period of 1/pq is exactly $\mbox{lcm}(r,s)$ and I'm done (I think). I'm a bit stuck with this last bit though. Am I on the right track here or is there some other obvious way of going about this?

Thanks for any help.