Determine the minimum value of a prime number P such that each of P-1 and P+1 has at least three distinct prime divisors.
Both P-1 and P+1 are multiples of 2 but only one is a multiple of 4. Halving both will leave one of them odd. so I visually scanned a table of https://oeis.org/A001221 looking for a 3 or higher next to a 2 or higher. Then I just doubled them both:
The first pair is 65 and 66
65=5*13 so 130=2*5*13
66=2*3*11 so 132=2^2*3*11
and so checking P=131 is prime
the next candidate pairs are (69,70) (77,78) (84,85) (90,91)
leading to primes 139 and 181
but 155 and 169 are composite
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Posted by Jer
on 2022-10-16 11:13:24 |
solution using OEIS
