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 $\sigma(A)$?
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8$\begingroup$ Bad idea to use $\sigma(A)$ for the number of prime factors of $A$ when 1) everyone else uses $\sigma$ for the sum of the divisors and 2) everyone else uses $\omega$ for the number of prime factors. $\endgroup$– Gerry MyersonCommented Dec 20, 2010 at 21:48
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$\begingroup$ If the q_i are all equal this should follow, at least modulo a few exceptions, from Zsigmondy's theorem: en.wikipedia.org/wiki/Zsigmondy's_theorem $\endgroup$– Qiaochu YuanCommented Jan 4, 2011 at 11:38
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$\begingroup$ @Qiaochu: You don't need them to be equal, you just need them to be powers of the same prime. Also Zsigmondy's theorem helps when $m_i$'s have lots of divisors. $\endgroup$– Gjergji ZaimiCommented Jan 4, 2011 at 12:04
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2 Answers
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No it's not true you have the following counterexample: $$ \frac{2^5-1}{2-1} \times \frac{5^3-1}{5-1} = 31^2 $$
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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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$\begingroup$ However, the question asks about $\omega(\sigma(B)) = \omega(mB)$, not $\omega(B)$. $\endgroup$ Commented Sep 4, 2011 at 5:43
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$\begingroup$ warn: Huan used
sigma'' to denoteomega''. Anyway what do you think is the interesting question here ?. No feedback seems known from the OP... $\endgroup$ Commented Sep 6, 2011 at 18:06