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Let $p$ be a prime of the form $4k+1$ , then can $\dfrac {p^p-1}{p-1}$ be a prime number ?

Peter
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  • @DirkLiebhold Did you verify this, or did you find a link ? – Peter Oct 17 '17 at 09:24
  • Regarding links, good idea @Peter: https://oeis.org/A001039 – Dirk Oct 17 '17 at 09:29
  • @DirkLiebhold PARI/GP is not the fastest program, but it takes already quite long for the range $[1,3000]$. The only explanation I have is that magma detected algebraic factors. – Peter Oct 17 '17 at 09:30
  • @Peter thanks for the hint, I walked right into a trap of magma as it seems (for details: in forming the fraction, magma sees the number no longer as an integer - although it is one - but as a rational number - primality tests for rational numbers are, of course, trivial...). I will check again with integers and come back with, most likely much smaller, numbers... – Dirk Oct 17 '17 at 09:40
  • 400 seconds for $[1,3000]$ @Peter , so magma is also rather slow. Sorry for my mistake... – Dirk Oct 17 '17 at 09:51
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    $\frac{p^p-1}{p-1} =1+p+p^2+\cdots +p^{p-1} = p(1+p(1+p(1+\cdots+p(1))))+1$ – Ahmad Oct 17 '17 at 10:04
  • Upto $p=5000$, the only primes would be $2,3,19,31$, but none of them has the form $4k+1$ – Peter Oct 17 '17 at 10:19
  • This article (https://number.subwiki.org/wiki/Congruence_condition_on_prime_divisor_of_cyclotomic_polynomial_evaluated_at_an_integer) states that all divisors of said number have to be congruent to $1$ mod $p$. – Dirk Oct 17 '17 at 10:25
  • https://math.stackexchange.com/questions/1997613/prove-that-fracpp-1p-1-is-not-prime-if-p-equiv-1-pmod-4 –  Oct 17 '17 at 10:25
  • @DirkLiebhold This is clear because the order of a prime factor $q$ modulo $p$ must be $p$. This gives better chances that such a number maybe prime. – Peter Oct 17 '17 at 10:28
  • According to my PFGW-calculation no such prime $p\le 6000$ exists, therefore such a prime must have more than $20\ 000$ digits. – Peter Oct 17 '17 at 10:41

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