What is the number of distinct primes $p$ such that $$\binom{\frac{p+1}2}2\ =\ 5\cdot r\cdot q$$ where $5<r<q<p$ are primes. (See the answer given by avz2611 in the following). Similarly, if $$\binom{\frac{p+1}2}2\ =2\cdot 3\cdot\ 5\cdot r\cdot q$$ what's the number of distinct primes $p$?

  • 1
    $\begingroup$ Please clarify. Do you mean that $r,q$ are fixed, and we're looking for $p$? Something else...? $\endgroup$ – Amitai Yuval Sep 27 '14 at 8:45
  • $\begingroup$ I am sorry that the question what I put is not clear. Here, $r,q$ are not fixed. I just want to find all the primes $p$ such that $\binom{\frac{p+1}2}2$ has three prime divisors $5,r,q$ with $5<r<q<p$. Please see the following. $\endgroup$ – Liu Sep 28 '14 at 1:21

$(p+1)(p-1)=40.r.q$ now (p+1) or (p-1) one of them have to be a multiple of 3 so you should have a multiple of 3 on the right hand side but 40 is not divisible by 3 and as both $r$ & $q$ are greater 3 and are primes even they are not divisible by 3 so no solutions possible

  • $\begingroup$ I'm very grateful to you for your help. And I think it is right. Thanks again. $\endgroup$ – Liu Sep 28 '14 at 0:58
  • $\begingroup$ Can I ask you a question more? If $(p+1)(p-1)=2^4.3.5.r.q$, then what about $p$? Do three many distinct primes $p$ exist? Thank you. $\endgroup$ – Liu Sep 28 '14 at 2:26

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.