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Old 2019-03-28, 13:45   #1
enzocreti
 
Mar 2018

21B16 Posts
Default triples of consecutive primes

(3^a)*(2^b)+1=p(n)*p(n+1)*p(n+2)


where a and b are non negative integers, and p(n),p(n+1),p(n+2) are consecutive primes.




A solution surely is a=1, b=7, p(n)=5, p(n+1)=7 and p(n+2)=11.


But with Pari i did not find any other solution.
Have you some explanation? The problem can be generalized?
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