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Prove that there are infinitely many prime numbers $p$ and $q$ such that $pq\mid2^{pq}-2$.

We may start with some easy steps. By Fermat's Little Theorem, $p\mid2^{p-1}-1$ and $q\mid2^{q-1}-1$.

According to my experience, it is very hard to "construct" prime numbers, so I suppose we may prove by contradiction. However, there are no trivial solutions, the smallest one is $(p,q)=(11,31)$.

Here are all solutions below $1000$.

{11,31}
{17,257}
{19,73}
{23,89}
{23,683}
{29,113}
{31,151}
{31,331}
{37,73}
{37,109}
{43,127}
{43,337}
{53,157}
{59,233}
{71,281}
{73,109}
{73,433}
{89,353}
{89,397}
{89,683}
{97,193}
{97,241}
{97,673}
{101,601}
{103,307}
{127,337}
{137,953}
{149,593}
{151,331}
{151,601}
{157,313}
{167,499}
{193,673}
{229,457}
{241,673}
{251,601}
{257,641}
{271,811}
{307,919}
{337,673}
{401,601}
{433,577}
TheJack
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youthdoo
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  • It is interesting that some prime numbers appear quite a lot of times while most never do. – youthdoo Aug 14 '23 at 10:01
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    Interestingly, the divisibility $p - 1 \mid q - 1$ appears quite often. Indeed, in this case $2^{q - 1} \equiv 1 \pmod{p}$ is always true. – Denis Shatrov Aug 14 '23 at 10:13
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    Composites $n$ satisfying $2^{n-1}\equiv 1\mod n$ are called Poulet-numbers. If $p,q$ are both odd this is equivalent to the desired divisibility. So, basically you search the semiprimes that are also Poulet-numbers. – Peter Aug 14 '23 at 10:16
  • By the way , it is easy to show that $p,q$ must both be odd to satisfy the given divisibility. – Peter Aug 14 '23 at 10:23
  • If $p$ is a Wieferich-prime , $p=q$ is a solution. – Peter Aug 14 '23 at 10:28
  • If we only look at the case where $p - 1 \mid q - 1$, then using the FLT multiple times we have that $pq \mid 2^{pq} - 2^p$, hence $pq \mid 2^{pq} - 2 \Leftrightarrow pq \mid 2^p - 2 \Leftrightarrow q \mid 2^{p-1} - 1$. Thus it would be sufficient to show that there's an infinity of couples $(p,q)$ such that $p-1 \mid q-1$ and $q \mid 2^{p-1} - 1$. In case that helps. – Bruno B Aug 14 '23 at 12:35
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    The first one $11*31=341$ is the first Fermat pseudoprime. You can compare your numbers to the list provided here (Poulet numbers...): https://en.wikipedia.org/wiki/Fermat_pseudoprime – user25406 Aug 14 '23 at 13:02

1 Answers1

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I believe this is what you're looking for: This is from a 2-page paper by Erdos titled "On the Converse of Fermat's Theorem": https://www.math.stonybrook.edu/~moira/mat331-spr10/papers/1949%20ErdosOn%20the%20Converse%20of%20Fermat's.pdf

Additionally, as pointed out by Peter in the comments, your question is equivalent to asking if there are infinitely many Poulet numbers with 2 factors (also super-Poulet since all divisors are prime) where a Poulet number is a number that is a pseudoprime to base 2. So by this definition, Erdos actually extends a proof by Lehmer to show there are infinitely many Poulet numbers with any number of factors. Not only that, his argument shows that, for any prime $p > 13 ~ $ there exists $ q $ for which $pq$ is a Poulet number.

TheJack
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