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2019-03-28, 13:45
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#1 |
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Mar 2018
10000110112 Posts |
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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