For any two distinct primes $p, q$ there is a unique integer $k$ such that: $$pq=\binom{k}{q}-\binom{p}{q}-\binom{k-p}{1}+1$$ Where $k$ is the smallest integer greater than $p$ that is relatively prime to $q$; I tested this for several primes and $k$'s satisfying the condition and I've found this working so far. I'll give one example:

Let $p$=$3$, $q$=$2$ and $k$=$5$ for $5$ is the smallest number greater than $3$ that relatively prime to $2$; therefore: $$3⋅2 = \binom{5}{2}-\binom{3}{2}-\binom{5-3}{1}+1$$$$⇒10-3-2+1$$$$⇒6$$

Sorry, that some of my questions have not been well-received and I'm in danger of being blocked from asking any more! I've read all the useful guidelines about asking good questions and topics and I tried my best to maintain all these according to my best understanding. Please guys go easy on me and I hope this time I post an on topic question. I really want to stay with MSI for a long time. I have no formal training on maths but I do love numbers and maths so I often play with them and try to find something out of it and share it. That's it, that's why I'm here.

I've tried to work on this problem by my self with my very limited knowledge. I realized that I'm not able to solve/prove it. I believe that would suffice a bit. If my tag(s) are wrong please add the right tag(s) for me, it will be much appreciated.

P.S.- Sorry, for my English, it's a second language.


  • $\begingroup$ What about $p = 13, q = 3$ ? It seems to be wrong. $\endgroup$ – dezdichado Aug 9 '14 at 21:15
  • $\begingroup$ Also k is p+1 or p+2 $\endgroup$ – Sandeep Silwal Aug 9 '14 at 21:32
  • $\begingroup$ Yes! I've just tested for it for $p$=$13$, $q$=$3$ and it doesn't hold. Thanks for your comment. @EnkhzayaEnkhttaivan $\endgroup$ – Tanin Aug 9 '14 at 21:34

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