Find $X$, $a$ : ALL prime factors of $(X^a - 1)/(X - 1) < X$ where $X$ is an odd prime, and $a$ is an odd integer.
For example, let $X = 37$, $a = 3$, we get $$\frac{37^3-1}{36} = 3 \times 7 \times 67.$$ When factoring numbers such as this, I've noticed that almost all have at least one prime factor larger than $X$ (e.g. 67 > 37). I would like to know for what values of $X$, $a$ are ALL of the prime factors of $(X^a-1)/(X-1)$ less than $X$. For example, let $X = 79$, $a = 3$, we get $$\frac{79^3-1}{78} = 3 \times 7^2 \times 43$$ and $43 < 79$.
My math education level is first year of high school so a transparent explanation, if possible, would be great. I understand basic congruences.
 A: (This is more a comment than an answer, but because I want to show images I'll put it in an answer-box) 
I've played around with this and considered the number of primefactors (with multiplicity) of $$ \small f(p,a) = {p^a - 1 \over p-1 } \qquad p \in \mathbb{P}$$
$$ \small  n(p,a) = \text{ number of primefactors in } f(p,a) $$
for $\small a=5,6,7 $ and $\small 1 \lt p \lt 200000 \qquad$ (I used $\small p \lt 20000 $ for $\small a=7$).         
The idea is, that if we have $\small n(p,a)<a$ , then at least one primefactor in $\small f(p,a)$ must be bigger than $\small p$.      
If the exponent a is prime, then we have "few" primefactors, and if the exponent a is composite then we'll have "many" primefactors. Here are three images of the frequency distributions for $\small a=5,6,7$ 
Distribution for a=5. For $\small n(p,5) \lt 5 $ (on the x-axis) we must have a primefactor $\small q \gt p$ :        

(source: helms-net.de) 
Distribution for a=7 . For $\small n(p,7) \lt 7 $ (on the x-axis) we must have a primefactor $\small q \gt p$ :        

(source: helms-net.de) 
Distribution for a=6:        

(source: helms-net.de) 
