My Math Forum Conjecture

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 March 12th, 2017, 04:01 AM #1 Banned Camp   Joined: Dec 2013 Posts: 1,117 Thanks: 41 Conjecture Conjecture about prime numbers : For every positive integer n>29 it always exist at least one pair of prime numbers p and q (included the number 2) such as : p*q = 1 mod n p and q both < n Example n=30 p=7 < 30 q=13 < 30 7*13=91=1 mod 30 There are only 3 numbers <30 (2,11,29) not satisfying the 2 conditions above. There is a general form of n satisfying both conditions : n=p^2-1 where p is prime Example : 8=(3*3)-1 3 < 8 9=1 mod 8 24=(5*5)-1 5<24 25=1 mod 24 etc.... Any proof or counterexample? Is there a way to test quickly any n to check if it has a solution?
 March 12th, 2017, 09:09 AM #2 Banned Camp   Joined: Dec 2013 Posts: 1,117 Thanks: 41 No counterexample yet? Maybe there is a proof (elementary?) that the conjecture is true.
 March 12th, 2017, 01:21 PM #3 Math Team     Joined: Jul 2011 From: North America, 42nd parallel Posts: 3,372 Thanks: 233 First I would like to say welcome back. I'm glad you are back. I will keep this interesting question in mind but my brain is mush right now. Thanks from mobel
 March 12th, 2017, 01:44 PM #4 Senior Member   Joined: Aug 2012 Posts: 1,960 Thanks: 547 Surely this would work for 11 and 29. To say $pq \equiv 1 \pmod n$ means that $p$ and $q$ are multiplicative inverses mod $n$. If $n$ is prime then everything other than $0$ or a multiple of $n$ is invertible in $\mathbb Z /n \mathbb Z$. For example $3$ and $4$ work mod $11$. In the general case you just have to find any invertible element in $\mathbb Z /n \mathbb Z$. It and its inverse will have representatives $< n$. An element is invertible if it's relatively prime to $n$. For example if $n = 100$ then all I have to do is find any number less than $100$ and relatively prime to it (other than $1$). For example $3$ would work. Then you just have to find the multiplicative inverse of $3$ mod $100$ and you're done. In this case $3$ doesn't divide $101$ but $3$ does divide $201$. Since $201 / 3 = 67$, your $p$ and $q$ are $3$ and $67$. Of course this solution is not unique, any pair of multiplicative inverses will do. To find a multiplicative inverse mod $n$ you can use the extended Euclidean algorithm. https://en.wikipedia.org/wiki/Extend...dean_algorithm Last edited by Maschke; March 12th, 2017 at 01:55 PM.
March 12th, 2017, 02:13 PM   #5
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Quote:
 Originally Posted by Maschke Surely this would work for 11 and 29. To say $pq \equiv 1 \pmod n$ means that $p$ and $q$ are multiplicative inverses mod $n$. If $n$ is prime then everything other than $0$ or a multiple of $n$ is invertible in $\mathbb Z /n \mathbb Z$. For example $3$ and $4$ work mod $11$. In the general case you just have to find any invertible element in $\mathbb Z /n \mathbb Z$. It and its inverse will have representatives $< n$. An element is invertible if it's relatively prime to $n$. For example if $n = 100$ then all I have to do is find any number less than $100$ and relatively prime to it (other than $1$). For example $3$ would work. Then you just have to find the multiplicative inverse of $3$ mod $100$ and you're done. In this case $3$ doesn't divide $101$ but $3$ does divide $201$. Since $201 / 3 = 67$, your $p$ and $q$ are $3$ and $67$. Of course this solution is not unique, any pair of multiplicative inverses will do. To find a multiplicative inverse mod $n$ you can use the extended Euclidean algorithm. https://en.wikipedia.org/wiki/Extend...dean_algorithm
It does not work for 11
3 is prime <11
4 is not prime
You need to read carefully my statement.
It does work for 2 and 29
Otherwise you need to prove that it works any n>29.
You did not until now.

 March 12th, 2017, 02:46 PM #6 Banned Camp   Joined: Dec 2013 Posts: 1,117 Thanks: 41 To find a multiplicative inverse mod n I will suggest you to read this article : https://f56df077-a-62cb3a1a-s-sites....attredirects=0 I implemented it quickly on Excel it works very fine.
 March 12th, 2017, 02:58 PM #7 Senior Member   Joined: Aug 2012 Posts: 1,960 Thanks: 547 Oh primes, sorry. Good question. Comes down to whether we can always find a pair of multiplicative inverses mod $n$ that are themselves prime. Sounds like a hard problem offhand. Last edited by Maschke; March 12th, 2017 at 03:06 PM.
 March 12th, 2017, 03:44 PM #8 Banned Camp   Joined: Dec 2013 Posts: 1,117 Thanks: 41 Both must be prime and less than n. So 2 conditions. Otherwise you will always find a solution with only one condition : both prime. It is linked to the factorization of odd semi-primes. m=91 (semi-prime to factorize) Hence n=90 If 90 has 2 primes p and q as factors such as p*q=1 mod 90 then 91 has 2 factors (7 and 13 for example). The important question remaining unanswered is how to find p and q quickly (that is another problem). The conjecture must be proven true first.
 March 18th, 2017, 03:44 PM #9 Senior Member   Joined: Aug 2008 From: Blacksburg VA USA Posts: 338 Thanks: 4 Math Focus: primes of course Perhaps this is obvious but since p,q

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