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The following question is equivalent to a problem in group theory. Let $ p > 13$ be a prime number distinct from 239. Let $ a=(p^2+1)/2 $. Is there any prime divisor $r$ of $a$ such that $r\mid a$ or $r^2\mid a$ and specially $ r^3 $ does not divide $a$ and also $(1+kr)\not\mid a$, for each nonzero $k$?

We check it for many primes as the computer allows us. Always we get the positive answer. Thanks for your answers.

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  • $\begingroup$ Are you saying that such $r$ exists for all $p$? $\endgroup$
    – joro
    Feb 27, 2014 at 13:29
  • $\begingroup$ Yes as we check almost we can find a prime number $r$. $\endgroup$
    – BHZ
    Feb 27, 2014 at 13:40

1 Answer 1

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If I understand correctly, there are primes $p$ for which $r$ with yours properties doesn't exist.

The smallest I found is $p=241727$ and $a=5 \cdot 13 \cdot 17 \cdot 29 \cdot 37 \cdot 41 \cdot 601$.

For each prime factor $r$ of $a$ there exist divisor $d$ of $a$ such that $d \equiv 1 \pmod{r}$, $d \ne 1$, which shows $1+kr \mid a$.

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  • $\begingroup$ If d is the number of prime divisors and $2^{d-1}$ is not far from the size of the largest prime divisor, the $(1+kr)$ condition seems likely to be satisfied. $\endgroup$ Feb 27, 2014 at 15:42
  • $\begingroup$ Dear Joro, Thank you very much for your help. Of course in group theory every group of odd order is solvable and the normal minimal subgroup of a solvable group is elementary abelian. Therefore there exists a Sylow subgroup which is normal and abelian which is equivalent to the above question. Therefore we can positively answer to the above question for this example using group theory methods. But here the order of each prime divisor is 1 and we could get this result. This statement is not applicable for the general case. $\endgroup$
    – BHZ
    Feb 27, 2014 at 23:07

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