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Let $k$ and $a_1,a_2 \cdots a_k$ be fixed integers, each of them being $>1$. Show that there are only finitely many $k$-tuples of primes $(p_1,p_2, \cdots p_k)$, with the following property: there exists a positive integer $m$ such that $$ \prod_{i=1}^k(a_i^{p_i}-1)=m! $$ This problem is from AMSP Cornell 2012. Could someone help me solve it? I tried to use Zsigmondy's theorem. But I don't know how to use the fact that the exponents are primes.

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    $\begingroup$ Can you give an example of $m$ for which there are infinitely many such tuples? $\endgroup$ – barak manos Jun 17 '14 at 13:02
  • $\begingroup$ Well no, I don't have any idea about it. This whole problem seems so unbelievable, I don't know how to start. Thanks $\endgroup$ – shadow10 Jun 17 '14 at 13:09
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For a positive integer $n$, let $v_3(n)$ denote the exponent of $3$ in the prime factorization of $n$. Let us fix some $k$-tuple of primes $(p_1,\ldots,p_k)$ and write $\displaystyle{R=\prod_{i=1}^k(a_i^{p_i}-1)}$. It is easy to see that $\displaystyle{v_3(R)=\sum_{i=1}^kv_3(a_i^{p_i}-1)}$. Clearly, if $3\vert a_i$ for any $i$, then $v_3(a_i^{p_i}-1)=0$. On the other hand, if $3\nmid a_i$, then the Lifting The Exponent Lemma tells us that $v_3(a_i^{p_i}-1)=v_3(a_i-1)+v_3(p_i)\leq v_3(a_i-1)+1$. Thus, $\displaystyle{v_3(R)\leq k+\sum_{i=1}^kv_3(a_i-1)}$, which puts an upper bound on $m$ that does not depend on our choice of the $k$-tuple of primes. From the upper bound on $m$, we get an upper bound on the possible primes $p_i$ that we may use in the $k$-tuple, so the number of such $k$-tuples must be finite.

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