2022-04-17, 17:28 | #1 |
"Daniel Jackson"
May 2011
14285714285714285714
2^{2}×3×59 Posts |
Primes p such that p^2+p+1=q is prime, and p is largest prime factor of q^3-1
I was wondering if there's even such a prime p that fits the following:
1. p^2+p+1=q is prime (seems that p must be of the form 6n-1, other than 2 and 3). 2. Largest prime factor of q^3-1 is p. So far, I haven't found any that fit #2 (searched up to p=997), but I'm guessing #1 has infinitely many solutions. Anyone know how I might do this in PARI/gp? I want to search up to at least p=9999991. |
2022-04-17, 18:03 | #2 |
Mar 2019
3·97 Posts |
How about:
Code:
? forprime(p = 2, 10000000, q=p*p+p+1; if(isprime(q) && vecmax(factorint(q^3-1)[, 1]) == p, print(p))) 125669 138209 254537 309629 532187 1107497 1126523 1210103 1225817 1287329 1524431 1534349 1539719 1720181 1793123 1814609 1861151 1920731 1932071 1974881 2270423 2366057 2490479 2494931 2530373 2586377 2725841 2755943 2782667 2885837 ... |
2022-04-18, 14:22 | #3 |
Feb 2017
Nowhere
2^{8}·23 Posts |
I note that, if p == 1 (mod 3) then q = p^2 + p + 1 is divisible by 3.
Also, q^3 - 1 = (q-1)*(q^2 + q + 1). If p > 2 then p is the largest prime factor of q-1 (proof: exercise). Also, q == 1 (mod p) so q^2 + q + 1 == 3 (mod p). So if p > 3 then p does not divide q^2 + q + 1, and we want the largest factor of q^2 + q + 1 to be less than p. Now N = q^2 + q + 1 is slightly larger than p^4, so we want the largest factor of N to be less than N^(1/4). The probability of a "random" number being that "smooth" is given by the "Dickman function" evaluated at 1/4, which is approximately .00491. |
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