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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 $p$?

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    $\begingroup$ 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
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$(p+1)(p-1)=40.r.q$ 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

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  • $\begingroup$ 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

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