If $5\mid\frac {n^{r-1}-1}{n-1}$ and $r $ is even then $r=10k+6$ Let $5\mid\frac {n^{r-1}-1}{n-1}$ and $r $ is even. I'm trying to show $r=10k+6$. 
Since 
$ \frac {n^{r-1}-1}{n-1}=n^{r-2}+...+n+1$ and I couldn't find any $n $ for which
$5|n^2+n+1$
$5|n^6+n^5+n^4+n^3+n^2+n+1$
$5|n^8+n^7+n^6+n^5+n^4+n^3+n^2+n+1$
$5|n^{10}+n^9+n^8+n^7+n^6+n^5+n^4+n^3+n^2+n+1$
But for $r=6=10\times0+6$ and $n=6$ we see that
$5|6^4+6^3+6^2+6+1=1555$

First of all, let's take $a=r-1$, since it is a bit easier to calculate with it.
Define $l$ like mentioned above $n^a-1=5l(n-1)$. Looking this equation with modulo $n$ we will get:
$$-1\equiv -5l \mod(n) \Leftrightarrow n \mid 5l-1$$
From this we get that 5 can not divide n, since otherwise we would have $5\mid 5l-1$, which is not possible. That means from Fermats Little Theorem we have $5|n^4-1$. If $5|n^a-1, \forall n\in \mathbb{N}$, then $a$ should be of the form $a=2t$ (except for $5\mid n-1$), which contradicts it being odd from your form of $a=r-1=10k+6-1=5(2k+1)$.
For $5\mid n-1$ it follows from the LTE lemma that $5\mid a$, but it can not be derived that $a$ has to be odd. Looking up on wolfram alpha also confirms this ex. $n=6, a=5,10,15,...$ works.
Where is this problem from?
If it really should hold that $a=5(2k+1)$, then we have two forms which need to hold and there can not exist a pair $(n,a)$ for which the above statement holds.
