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$\DeclareMathOperator{\Ord}{Ord}$When does $q^b = (p^a - 1)/(p - 1)$ for $p, q$ odd primes and $a, b$ odd integers $> 1$ ? If no examples are possible, please give a simple proof.

A proof of this for $q=3, p=5$ might be

Assume $3^b = (5^a - 1)/4$; then \begin{equation} 5^a - 4 \cdot 3^b = 1. \tag{1} \end{equation} We show that the exponents $a,b$ in (1) are even.

We see that $3^b \mid 5^a - 1$ $\Rightarrow$ $5^a \equiv 1 \bmod 3$. Also $5 \mid 4\cdot3^b+1$ or $4 \cdot 3^b \equiv -1 \bmod 5$ so that $3^b \equiv 1 \bmod 5$.

In short: \[ 3^b\equiv1 \bmod 5, \quad 5^a\equiv1 \bmod 3. \]

By a well known theorem: if $X^c \equiv 1 \bmod p$, then $\Ord(c,p) \mid c$. By inspection, $\Ord(3,5) = 4$ so that $4 \mid b$. Similarly, $\Ord(5,3) = 2$ so that $2 \mid a$.

This violates our assumption that $a,b$ are odd. Proof complete.

However, moving further:

As $a,b$ are both even we can write (1) as $1 = 5^{2A} - 4\cdot 3^{2B} = (5 - 2\cdot3^B)(5 + 2\cdot3^B)$. Thus, $5 + 2\cdot3^B = 1$ but there is no positive value of $B$ satisfying this. Therefore (1) is not true.

Note: I want to be clear that the above proof, if it is correct, is not mine but was given to me by someone I am not free to name. If it is incorrect, the fault is entirely mine.

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    Nick, please don't stop posting. It's better to have the question unLaTeXed, than to have no question at all. There are loads of people around here who will be happy to tidy up the formatting of your questions, as Dylan did on this occasion.2012-06-20

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Your question is a specialization of this question and this MO question linked there. So it is almost certain that no example (with $b>1$) is known, nor any simple proof that none is possible.

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Well, there are examples, say $\frac{5^3-1}{5-1}=31^1=31$or $\frac{7^5-1}{7-1}=2801^1=2801$

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    Good point. I meant a,b > 1 and have edited my question to reflect that. Thanks for your feedback2012-06-20
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q^b = (p^a - 1)/(p - 1)

Now for q^b to be defined p is ne 1

p is an odd prime number therefore p = 3 or p = (6k - 1) or p = (6k + 1) where k is a positive integer.

when p = 3

q^b = (3^a - 1)/(3 - 1)

when p = 6k - 1

q^b = ((6k - 1)^a - 1)/(6k - 2)

= [(6k - 1 - 1)(6n + 1)]/(6k - 2) in atleast one of the instances.  = [(6k - 2)(6n + 1)]/(6k - 2)  = 6n + 1 where n is a positive integer 

when p = 6k + 1

q^b = ((6k + 1)^a - 1)/(6k)

= [(6k + 1 - 1)(6N - 1)]/(6k) in atleast one of the instances.  = [(6k)(6N - 1)]/(6k)  = 6N - 1 where N is a positive integer 
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    (7^5 - 1)/(7 - 1) = (7^4 + 7^3 + 7^2 + 7 + 1) = 2801 = ((6)(467) - 1)2012-07-12