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$\sum_{k=0}^{n-1}(5^k)$ divide 13 only if $5^n-1$ divide 13, with n in $\mathbb{N*}$

I proved that $5^n$ is prime with 13 (n in $\mathbb{N}$)

we have: 13 divide $5^n$-1 $\Rightarrow (13 * \sum_{k=0}^{n-1}(5^k)$) divide $(5^n-1) * \sum_{k=0}^{n-1}(5^k)$

I tried to use Gauss lemma 13 divide $(5^n-1) * \sum_{k=0}^{n-1}(5^k)$ and 13 is prime with $5^n - 1$

$\Rightarrow$ 13 divide $\sum_{k=0}^{n-1}(5^k)$

the problem I dont have 13 is prime with ($5^n - 1$); I only have 13 is prime with $5^n$

Malek Gara-Hellal
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  • $5^n$ never divides $13$. Well, unless you are allowing $n=0$. – lulu May 16 '21 at 18:56
  • @lulu i meant $5^n−1$ divide 13. sorry, not $5^n$ divide 13 – Malek Gara-Hellal May 16 '21 at 18:59
  • Well, I think you meant $13$ divides $(5^n-1)$, right? – lulu May 16 '21 at 18:59
  • Yes! exactly that's what I meant – Malek Gara-Hellal May 16 '21 at 19:00
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    @MalekGara-Hellal Hint: The summation is of a geometric series. – John Omielan May 16 '21 at 19:01
  • Ok, so we want $5^n\equiv 1 \pmod {13}$. But that's easily solved, and the least such $n$ is $4$. – lulu May 16 '21 at 19:01
  • @JohnOmielan Thank you! but I didn't really seem to get that hint I know that $\sum_{k=0}^{n-1}(5^k) = \frac{(1-q^n)}{1-q}$ I didn't get the relation between them – Malek Gara-Hellal May 16 '21 at 19:07
  • @lulu, my problem is how to prove that that summation is dividable by 13 not the $5^n-1$ – Malek Gara-Hellal May 16 '21 at 19:10
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    @MalekGara-Hellal You're welcome. Note the summation formula gives $\sum_{k=0}^{n-1}(5^k) = \frac{5^{n} - 1}{4}$. Thus, if $13 \mid \frac{5^{n} - 1}{4}$, what does that say about whether or not $13$ divides $5^{n} - 1$ (i.e., whether or not $5^{n} - 1$ is divisible by $13$)? – John Omielan May 16 '21 at 19:12
  • @JohnOmielan Oh I got you now! but isn't it wrong to multiple by 1/4? its not an Integer. – Malek Gara-Hellal May 16 '21 at 19:14
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    Since $\gcd (13,4)=1$, that shouldn't be a problem. – BR Pahari May 16 '21 at 19:15
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    @MalekGara-Hellal All the division by $4$ means, since we know that $\frac{5^n-1}{4}$ is an integer, is that $4$ is a factor of $5^n - 1$. It doesn't affect whether or not $13$ divides $\frac{5^n - 1}{4}$ (and also $5^n - 1$) because, as Basanta Raj Pahari just commented, we have that $\gcd(4, 13) = 1$. – John Omielan May 16 '21 at 19:16
  • Thank you very much! I really appreciate the help <3

    Just another question, let's call that summation Un. $Un = \frac{(5^n-1)}{4}$, is this correct? I want to make sure from that so I can use Gauss Lemma: We have 13|($5^n-1$) then we have 13|4($5^n-1$) --> 13|(4Un). Using Gauss Lemme 13|(4*Un), 13 and 4 are prime between each other. -->then we have 13|Un

     – Malek Gara-Hellal May 16 '21 at 19:21

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