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#1 |
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Mar 2018
2×5×53 Posts |
The prime 5101 is congruent to 3^6 (mod 1093)
Are there infinitely many primes p congruent to 3^k (mod 1093) with k>0? |
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#2 |
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Aug 2006
3·1,993 Posts |
For any positive integer k there are infinitely many primes congruent to 3^k mod 1093, yes. Note that 3 is relatively prime to 1093 (and that 3^k has order 7 mod 1093).
Last fiddled with by CRGreathouse on 2019-04-15 at 17:02 |
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#3 | |
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Mar 2018
2×5×53 Posts |
Quote:
51001 divides 2^(5101-1)-1. Just a curio But maybe a better curio is that 1021 and 1201 divide 2^(5101-1)-1 where 1021 and 1201 are primes with 0 and 2 changed....1021+1201 is 2222...moreover 1801 and 8101 divide 2^5100-1...also here 8 and 1 changed Last fiddled with by enzocreti on 2019-04-15 at 17:52 |
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#4 |
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Mar 2018
21216 Posts |
(2^5100-1)/1021/1201/1801/8101 is congruent to 9393 mod (101^2)
(2^5100-1) is divisible by at least four primes of the form x^2+45*y^2 Last fiddled with by enzocreti on 2019-04-15 at 18:28 |
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