Prove $\frac{2^{ab}-1}{2^a-1} $ is composite number let $b>a>1$ be postive integers,show that
$$\dfrac{2^{ab}-1}{2^a-1} $$ is composite number
I try use $$(x^n-1)=(x-1)(x^{n-1}+x^{n-2}+\cdots+x+1)$$
so $x-1|x^n-1$.then let $x=2^a,n=b$ we have
$$\dfrac{2^{ab}-1}{2^a-1}=2^{(b-1)a}+2^{(b-2)a}+\cdots+2^a+1$$
but How to prove $2^{(b-1)a}+2^{(b-2)a}+\cdots+2^a+1$ is composite ？
 A: You also need $a >1$.
There is another less direct way to do this. Let us write $A=2^a-1$, $B=2^b-1$, and $C=2^{ab}-1$. Then both $A$ and $B$ divide $C$ so write $C=BD$ for some integer $D$. Also, the inequality $a <b$ gives the inequality $A<B$. Now, $a \ge 2$ also gives
$$B^2 < 2^{2b} -1$$ $$\le 2^{ab}-1 = C,$$ so in particular $B^2< C$, so $D$ must satisfy $D >B$ and [as $A < B$] in particular $A < D$.
Now suppose $p=\frac{C}{A}=\frac{BD}{A}$ for some prime $p$. We show next that this is impossible. First note that for $\frac{C}{A}=\frac{BD}{A}=p$ with $p$ prime, to be true, the prime $p$ must divide either $B$ or $D$.
Case 1: If $p$ divides $B$ then $B \ge p$ and as $D \ge B > A$, it follows then that $\frac{D}{A}>1$, and thus:
$$\frac{C}{A} = \frac{BD}{A} \ge \left(p × \frac{D}{A}\right) > p,$$ and in particular $\frac{C}{A} > p$, so we arrive at a contradiction.
Case 2: If $p$ divides $D$ then $D \ge p$, and as $B > A$, it follows then that $\frac{B}{A} > 1$, and thus:
$$\frac{C}{A} = \frac{BD}{A} \ge p ×\frac{B}{A} > p,$$ and in particular $\frac{C}{A} > p$, so we arrive here at a contradiction as well.
Note that this breaks down if $a=b$ [Case 2 in particular], or if $a=1$ [in which case $A=1$ and $D=1$]. Indeed, try $a=3$, $b=3$, and then try $a=1$, $b=5$. So you do indeed need $a<b$ and $a >1$.

You could do this another way. Use $C=BD$, with $A<\min\{B,D\}$, as established above. Then that $\frac{C}{A}=\frac{BD}{A}$ is an integer means $A$ can be factored $A=A_1A_2$ with $A_1|B$ and $A_2|D$ [one of $A_1,A_2$ may be $1$ and the other $A$]. Then $B$ can be written $B=A_1B'$, and furthermore, since $B>A \ge A_1$, it follows that $B' > 1$. Likewise, $D$ can be written $D=A_2D'$, and since $D>A \ge A_2$, it follows that $D'>1$.
Then:
$$\frac{C}{A}=\frac{BD}{A}$$ $$= \frac{(A_1B')(A_2D')}{A_1A_2} = B'D'.$$
As $B'$ and $D'$ are both integers greater than $1$ gives $B'D'$ composite, and thus, $\frac{C}{A} =$ $\frac{BD}{A}= $ $B'D'$ composite.
A: We use this $fact*$ that:
if m is a natural number and $a>1$ we have:
$$\big(\frac{a^m-1}{a-1}, a-1\big)= (a-1, m)$$
Proof:
Suppose $\big(\frac{a^m-1}{a-1}, a-1\big)=d$
Using following identity:
$\frac{a^m-1}{a-1}=(x^{m-1}-1)+(x^{m-2}-1)+(x^{m-3}-1)+\cdot\cdot\cdot+(a-1)+m$ $\space\space\space\space\space (1)$
and the fact that ($a^k-1$) is divisible by $(a-1)$ we conclude that m is divisible by d. If numbers $(a-1)$ and $m$ have a common divisor like ($\sigma>d$), considering relation (1) we conclude that ($\frac{a^m-1}{a-1}$) is divisible by ($\sigma$) and numbers ($\frac{a^m-1}{a-1}$) and ($a-1$) have a common divisor $(\sigma>d)$., which results in $(a-1, m)=d$
Now we have :
$A=\frac {2^{ab}-1}{2^a-1}=\big(\frac{(2^a)^b-1}{2^a-1}\big)$ which has a common divisor like d with $(m=b)$ that means $A$ is divisible by d, so it is a composite number.

*

*This fact and its proof is from a number theory book by Sierpinski.

