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Here is the original theorem from my book (A Course in Number Theory by H.E.Rose, 2nd edition):

Let $a>1$ and $c>1$ be integers. The integer $a^c-1$ is composite if $a>2$ or if $c$ is composite.

I am having trouble understanding part of the proof for this theorem. I understand that if $c$ is composite then $a^c-1$ is also composite. My book, however, does not provide much explanation for why $a^c-1$ is composite if $a>2$. I tried working it out myself and have come to the conclusion that if $a>2$ and $a$ is odd, then $a^c-1$ is even and therefore must be composite. I'm stuck with the case for $a>2$ and $a$ is even.

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  • $\begingroup$ Sometimes it helps to try out some values. Let's take $a = 10$ for a convenient decimal representation. We can write $10^n - 1$ as a sequence of $n$ 9's. Of course, this sequence is divisable by 9. The same technique works in any base a > 2. $\endgroup$ – Ruben May 16 '14 at 17:33
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Hint: $a^c-1=(a-1)(a^{c-1}+a^{c-2}+\dots+a^2+a+1)$

If $a>2$, then $a-1>1$. If $a=2$ and $c=mn$ with $m,n>1$, then $$ a^{mn}-1=b^n-1=(b-1)(b^{n-1}+\dots+b+1) $$ where $b=a^m$.

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We have

$$a^c-1=(a-1)\sum_{k=0}^{c-1}a^k$$ and if $a>2$ then $a-1\ne1$ and $\sum\limits_{k=0}^{c-1}a^k\ge1+a$. Conclude.

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You always have $a^c-1=(a-1)(a^{c-1}+a^{c-2}+\cdots+a+1)$, if $a>2$ and $c>1$ this is a proper factorization of $a^c-1$.

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