I need to prove that there is a prime number $q$ which divides the numerator of $\frac{x^{\ell e}-1}{(x^{\ell}-1)(x^{e}-1)}$ and doesn't divide the denominator. Here, $\ell,e,x$ are integers, and $x$ is of the form $p_1^{p_2^{\ell-1}}\cdot p_2^{p_1^{e-1}}$, where $p_1,p_2$ are primes distinct from $q$, and $\ell =p_1$, $e=p_2$.
- Forgot something important: $\ell e>\ell+e$.
Attempt:
I have tried using the next sentence to make it easier for me:
$$(x^{\ell}-1,x^{e}-1)=x^{(\ell,e)}-1$$ Since $(\ell,e)=1,$ we get: $(x^{\ell}-1,x^{e}-1)=x^{(\ell,e)}-1=x-1$.
To prove that there is a prime $q$ divides only the numerator, we shall check whether the number $\frac{x^{\ell e}-1}{(x^{\ell}-1)(x^{e}-1)}$ is integer, and after division of the numerator by the denominator, there is $q$ in the factorization.
Without any success from here.
Best Answer
We can write:
$$\frac{N=(x^l)^e-1}{D=(x^l-1)(x^e-1)}=\frac{(x^l)^{e-1}+(x^l)^{e-2}+\cdot \cdot \cdot +1}{x^e-1}$$
If $e=2k$ is even then:
$x^l+1|(x^l)^{e-1}+(x^l)^{e-2}+\cdot \cdot \cdot +1$
Therefore:
$x^l+1|N$
Also we have:
$D=x^e-1=(x^k-1)(x^k+1)$
If $l\neq k=\frac e2$ then $x^l+1$ divides N but does not divide D.So $x^l+1$ must be primes .For example for $l=2$ and $x=2$ gives $q=5$ so if e=6 , for instance, we get:
$D=(2^2-1)(2^6-1)=189$
$N=2^{12}-1=4095$.
The fraction is symmetric for and l and e so we may change e with l in above argument; the condition is:
$L\neq \frac e2$
Or:
$e\neq \frac l2$
That is one of l or e must be even.
Update:
Suppose e and l are odd, then it is possible to have a prime satisfying the condition for example for x=2, l=3, e=5 we get:
$\frac ND=\frac{2^{15}-1}{(2^3-1)(2^5-1)}=\frac{151}{7\cdot 31}$
So prime we are looking for is $151$.
Also for certain values of x, e and l, the numerator my be divisible by denominator. The condition must be discussed in separate question. For example for x=2,l=2, e=5 we get:
$\frac ND=31$
So 31 divides numerator.
Or with x=3, l=2 and e=5 we get:
$\frac ND=4 \times 61$
That means prime 61 only divides the numerator.