Belgian mathematician Catalan in $1876$ made next conjecture: If we consider the following sequence of Mersenne prime numbers: $2^2-1=3 , 2^3-1=7 , 2^7-1=127 , 2^{127}-1$ then $$2^{2^{127}-1}-1$$ is also prime number. The last term has more than $10^{38}$digits and cannot be tested at present, so I would like to know is there any theoretical indication that Catalan's conjecture could be true ?


At London Curt Noll's prime page I have found statement that this number has no prime divisors below $5*10^{51}.$

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    $\begingroup$ Note that this is not the same conjecture of Catalan proven by Preda Mihăilescu in 2002. $\endgroup$ Oct 11, 2011 at 10:28
  • $\begingroup$ I've never seen any argument in its favor. $\endgroup$ Oct 11, 2011 at 11:25
  • $\begingroup$ @Dan: Hopefully the new title is less confusing on that front. $\endgroup$ Oct 11, 2011 at 11:32
  • $\begingroup$ @Gerry,There are well known Cunningham chains...so this sequence might be chain of some specific length size... $\endgroup$
    – Peđa
    Oct 11, 2011 at 11:40
  • $\begingroup$ math.sjsu.edu/~hsu/courses/126/Law-of-Small-Numbers.pdf $\endgroup$
    – Peđa
    Oct 29, 2011 at 19:29

2 Answers 2


I don't know of any theoretical reason to think it would be true. I would conjecture it to be false.

Standard heuristics suggest that the first unknown example would be prime with probability $$e^\gamma\cdot2^{-120}\approx1.34\cdot10^{-34}\%$$ which is small.

  • $\begingroup$ I downvoted this answer. I see this answer as smart alik. The author probably already knows that a normal intuitive probabilistic argument generates that kind of result. It appears that they were looking to see whether it's possible that all the numbers in the sequence are prime with an explanation. $\endgroup$
    – Timothy
    Apr 5, 2020 at 2:11
  • $\begingroup$ @Timothy I haven’t seen any mathematicians suggest a reason it would be true. I believe my answer correctly summarizes the current state of knowledge on the conjecture. If you have any additional knowledge, please share! Tao has a metric of complexity described in his paper “Linear equations in primes” in which he presents solutions to certain problems of finite complexity; the twin prime conjecture is of infinite complexity. This problem is far beyond what can even be described by that measure. $\endgroup$
    – Charles
    Apr 5, 2020 at 18:19
  • $\begingroup$ It's okay to sometimes write an answer that gets downvoted. Shortly after I saw the answer, I realized that maybe smart alik wasn't your intent and it just appeared that way. I'm not sure what I was thinking. Different people have different opinions. There is a general tendency for people to induce bias in their opinion of things that personally affect them. There is no way to please everyone. I think it's better to go by how the answer appears and not by your intent. This answer still doesn't appear to answer the question. The question was about whether it could be the case that all the $\endgroup$
    – Timothy
    Apr 6, 2020 at 0:45
  • $\begingroup$ members of the sequence are prime with an explanation. Downvotes are meant for useless answers that cannot be fixed up to not be useless without rewriting them entirely. I think an answer that doesn't answer the question is useless. $\endgroup$
    – Timothy
    Apr 6, 2020 at 0:46

It's possible that all numbers in that sequence are prime and there is an explanation. Bertrand's postulate states that there is no prime number that is more than double the previous prime number. It may seem like there's no possible way to prove it. It turns out that a not very long proof of Bertrand's postulate can be found in the Wikipedia article Proof of Bertrand's postulate. It derived some statements that don't appear to be useful but in the end, they all got combined to derive the final result.

The fact that $2^{2^{127} - 1} - 1$ has no prime divisors below $5 \times 10^{51}$ is not a very hopeful sign at all that $2^{2^{127} - 1} -1$ is prime. I believe that given that information, there is a 50% chance of there being a factor below $25 \times 10^{100}$. If one isn't found below $25 \times 10^{100}$, there's still a 50% chance that one will be found below $625 \times 10^{200}$. How many times do you have to square that in order to get past the square root of $2^{2^{127} - 1} - 1$?

However, people already managed to prove that $2^{82,589,933} - 1$ is prime. Maybe some day, they will prove that $2^{2^{127} - 1} - 1$ is prime.

I've noticed a pattern; 31, 89, and 127 are all Mersenne prime exponents. 31 and 127 are themselves Mersenne primes. 89 is a factor of $2^{11} - 1$.

For any positive integers, $x$, $y$, and $z$, we say that $x$ is congruent to $y$ modulo $z$ when the difference between $x$ and $y$ is a multiple of $z$. That statement is written $x \equiv y \mod z$. A group is an ordered pair of a set and a binary operation on that set, $(S, \cdot)$ that satisfies the following properties.

  • $\cdot$ is associative
  • There is an identity element $e$
  • Every element $x$ has an inverse $x^{-1}$ which means $x \cdot x^{-1} = e$ and $x^{-1} \cdot x = e$

It is a well known theorem that the multiplication modulo group of any prime number is cyclic. This shows that for any prime number $p$, all factors of $2^p - 1$ must be of the form $2kp + 1$. However, when a prime number is of the form $2kp + 1$, it's probability of being a factor of $2^p - 1$ is far higher than you would normally expect.

Maybe just like we could form many results and combine them into a proof of Bertrand's postulate, we can do a similar thing for proving that all numbers in the sequence you listed are prime.

There is a not very hard to prove already known theorem that for any odd prime number $p$, 2 is a square modulo $p$ if and only if $p$ is congruent to 1 or 7 modulo 8. I've noticed by observation that there is a pretty nice and simple way of determining of determining when ever any positive integer is a square modulo any prime number but failed to figure out how to prove it. It's possible that there is another simple test for determining for any larger prime number than 3 that's congruent to 1 modulo 6 whether 2 is a cube modulo that number. Maybe there's some special property of numbers in the sequence you listed. Maybe there is a nice easy way to determine when a number is a cube modulo any prime number congruent to 1 modulo 6, a fifth power modulo any prime number congruent to 1 modulo 10.

Also take the sequence $s_0 = 4$, $s_1 = 4^2 - 2 = 14$, $s_2 = 14^2 - 2 = 194$, $s_3 = 194^2 - 2 = 37634$ and so on. According to the Wikipedia article Lucas-Lehmer primality test, for any odd prime $p$, $2^p - 1$ is prime if and only if $s_{p - 2}$ is a multiple of $2^p - 1$. Maybe we could also derive other statements and combine them with those statements to derive the final result that all numbers in the sequence are prime.


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