# When does $q^b = (p^a - 1)/(p - 1)$?

$\DeclareMathOperator{\Ord}{Ord}$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 $$5^a - 4 \cdot 3^b = 1. \tag{1}$$ 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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I TeX'ed the question because I wanted to read it. If you plan to ask more questions here, it would be worth learning the (basic!) commands involved in typesetting something like this. Take a look at the new source, and let me know if you have any questions. – Dylan Moreland Jun 20 '12 at 0:06
Mr. Moreland. I appreciate your effort very much. It looks GREAT! I have tried to LaTex my questions but it ends up totally scrambled. It may be due to my Web browser which is very restricted by my work. Or, it could just be incompetence on my part. You are not the first to complain so I will probably just stop bothering folks with my posts. I'm not very good at math anyway. Just curious about it. Again. thanks so much – Nick Jun 20 '12 at 0:14
@Nick I can't speak for everyone, but I'd rather that you continued to post your questions. – Dylan Moreland Jun 20 '12 at 1:13
@Nick: I forgot to mention this in the above comment, but I too would love to see more of your questions. At least to me, they seem quite interesting. – Dejan Govc Jun 20 '12 at 1:18
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. – user22805 Jun 20 '12 at 9:08

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 feedback – Nick Jun 20 '12 at 9:24

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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A careful observation reveals that none of the equation holds for odd number b > 1 – Rajesh K Singh Jul 12 '12 at 9:23
For all other instances (p^a - 1)/(p - 1) is a product of two prime numbers. – Rajesh K Singh Jul 12 '12 at 11:00
(7^3 - 1)/(7 - 1) = (7^2 + 7^1 + 1) = (49 + 7 + 1) = 57 = (3)(19) – Rajesh K Singh Jul 12 '12 at 11:02
(5^5 - 1)/(5 - 1) = (5^4 + 5^3 + 5^2 + 5^1 + 1) = (625 + 125 + 25 + 5 + 1) = (781) = (11)(71) – Rajesh K Singh Jul 12 '12 at 11:18
N is a positive integer ge 467 – Rajesh K Singh Jul 12 '12 at 12:32