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If $a$ is not a multiple of a prime $p$ , then prove that there is an integer $b$ such that $p^b-1$ is a multiple of $a$

I have no idea where to start I would be grateful if anyone can give me hint.

NOTE: Please try to prove it using only basic properties of GCD and LCM .Please avoid fancy theorems because I am not that advanced

Bill Dubuque
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Mathematical Curiosity
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    Avoid asking questions stating [I have no clue](https://math.meta.stackexchange.com/questions/26285/homework-reasonable-to-have-no-clue) . Show us your efforts so we can write more appropriate question . Also read this [How to ask good questions.](https://math.meta.stackexchange.com/questions/9959/how-to-ask-a-good-question) – The Demonix _ Hermit Dec 01 '19 at 12:01
  • Hint: there are only finitely many different possible remainders resulting from division by $a$. So there must be $b,c$ such that $p^b-1$ and $p^c-1$ leave the same remainder upon division by $a$. What does that say about the remainder of $p^b-p^c$ when divided by $a$? But $p^b-p^c=p^r(p^s-1)$ for appropriate $r,s$. – Gerry Myerson Dec 01 '19 at 12:14
  • Thanks,that was helpful.But how you are able to get that idea?.At starting it seemed as if it was unrelated but later on it popped the proof like magic.Any tips on thought process?? – Mathematical Curiosity Dec 01 '19 at 12:29
  • Who are you referring to ? – The Demonix _ Hermit Dec 01 '19 at 12:34
  • Sorry for not mentioning.My comment is actually meant for Gerry Myerson – Mathematical Curiosity Dec 01 '19 at 12:42
  • Then refer to @B.Goddard answer – The Demonix _ Hermit Dec 01 '19 at 12:43
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    Myerson and I live on opposite points on the earth. Why are we both awake at the same time? – B. Goddard Dec 01 '19 at 13:11
  • [Hint](https://math.stackexchange.com/a/61009/242) $\rm\ \overbrace{|\Bbb Z_a|<\infty\ \Rightarrow\bmod a\!:\ p^j\equiv p^k}^{\rm pigeonhole},\: j – Bill Dubuque Dec 01 '19 at 15:54
  • @B.G you've taken up residence in the middle of the Atlantic Ocean? – Gerry Myerson Dec 01 '19 at 21:38
  • If you want to be sure, Mathematical Curiosity, that I see a comment intended for me, you have to include @Gerry in it. Anyway, there was no thought process, just decades of experience. – Gerry Myerson Dec 01 '19 at 21:44
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    Maybe you would like to "accept" one of the answers, Mathematical, by clicking in the check mark next to it, if you have found that it gives you what you need. – Gerry Myerson Dec 03 '19 at 12:07
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    @GerryMyerson It seems that the O.P has never accepted any answers in all his questions on Math SE. – The Demonix _ Hermit Dec 03 '19 at 12:14
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    Is that true, Mathematical? If so, it would be really nice if you could go back over your other questions to see whether you'd like to accept answers to some or all of them. – Gerry Myerson Dec 03 '19 at 12:24

2 Answers2

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The set of numbers $\{p^x\}$ for $x$ a natural number, is infinite. If you divide then all by $a$ and look at the remainders, there are only $a$ possibilities. Therefore at least two of the numbers have the same remainder when you divide by $a$. (Pigeonhole principle.)

That leads to $p^x - p^y$ being a multiple of $a$ for some $x$ and $y$. Now you have a clue.

B. Goddard
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If $a$ is not a multiple of $p$ , we can write :

$$p\equiv r \mod a \quad\quad \text{such that }r \ne 0$$

We use Euler's theorem , to say that :

$$p^{\phi(a)} \equiv 1 \mod a$$

This follows from the fact that $p$ is a prime and $a$ is not a multiple of $p.$Hence it is a necessary condition for the question to remain true. Now taking $\phi (n) = b$ , we get :

$$p^b \equiv1\mod a \implies \boxed{p^b-1 \equiv \mod a}$$

Hence there always exist a multiple $b$ , for which $p^b-1$ is a multiple of $a.$

The Demonix _ Hermit
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  • $\phi(a)$ not $\phi(n)$ –  Dec 01 '19 at 14:56
  • Fermat only considered the case where $a$ is prime. The generalization you cite is due to Euler. Also, OP might consider Euler's result to be a "fancy theorem" going well beyond "basic properties of gcd and lcm". – Gerry Myerson Dec 01 '19 at 21:42