Let $F(n) = \underbrace{111..11}_{n \text{times}}$
Proof that if $p|F(n)$ then $gcd(n, p-1) > 1$
(p - prime and $p>3$)

My approach

If $n$ is even it is true because $p-1$ is even too so $$gcd(n, p-1) \ge 2 > 1 $$

But I completely don't know how to start with odd numbers...


$n=5$ $F(n) = 11111 = 271\cdot 41$ $gcd(5,270) > 0$ $gcd(5,40) > 0$


From answer I know that $$ 10^{n}\equiv 10^{p-1}\equiv 1\mod p $$ so $$gcd(n,p-1) = gcd(log_{10}(pk+1), log_{10}(pt+1)) \text{ for some k,t} $$ I am trying to show that it implies thesis...


$9F(n)=10^n-1$, so if $p\mid F(n)$ then $10^n\equiv 1\pmod p$, that is, $n$ is a multiple of the order of $10$ in $\Bbb Z_p^\times$, which is a non trivial divisor of $p-1$.

  • 1
    $\begingroup$ Just to put it in other words: $10^{n}\equiv 10^{p-1}\equiv 1\mod p$ So, $ord_p(10)$ divides $(n,p-1)$. $\endgroup$ – Julian Mejia May 14 at 17:04
  • $\begingroup$ what is $ord_p(10)$? I didn't have orders and groups on my lecture :( $\endgroup$ – Witalij May 14 at 17:16
  • $\begingroup$ and how you got $10^{p-1}\equiv 1\mod p$? $\endgroup$ – Witalij May 14 at 17:30
  • $\begingroup$ @Witalij If $p$ is neither $2$ or $5$, then $10^{p-1} \equiv 1 \pmod p$ due to Fermat's little theorem. $\endgroup$ – John Omielan May 15 at 2:47
  • $\begingroup$ Can it be solved without informations about $ord_p(10)$? $\endgroup$ – Witalij May 15 at 10:06

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