Question

Let $m_i\geq 2 (1\leq i\leq n)$ be $n$ pairwisely coprime positive integers and let $q_i\geq 2 (1\leq i\leq n)$ be $n$ arbitrary prime powers, let$A=\prod_{i=1}^n(({q_i}^{m_i}-1)/(q_i-1))$. Let $\sigma(A)$ be the number of different prime factors of A, is it true that $\sigma(A)\geq n$? If this is not true, is there a counterexample? Is there a good way to estimate $\sigma(A)$?

Answer 1

No it's not true you have the following counterexample: $$ \frac{2^5-1}{2-1} \times \frac{5^3-1}{5-1} = 31^2 $$

Answer 2

Let take a look to the special case when your $q_i$ are actually $n$ distinct \emph{odd} prime numbers.

I use the standard notations : $\omega(H)$ is the number of distinct prime divisors of $H$ and $\sigma(G)$ is the sum of all positive divisors of $G.$

Put $B$ the product of all the $q_i^{m_i}$

then we have

$$ \sigma(B) = mB $$

if $B$ is an odd $m$-multi-perfect number.

((sure, we do not known concrete examples of this, but...)

So, in this case

$$ \omega(B) = n $$

and you have your lower bound attained.

luis

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