Triple Mersenne Primes Proof where $p$ is either prime or non-prime and On the Matter of Triple Mersenne Numbers

$M_{M_{M_{p}}} = {2^2}^{2^{p - 1}} - 1, p \geq 1$ and either a prime or a non-prime

$p = 1, {2^2}^{2^{1 - 1}} - 1 = 4 - 1 = 3$ is a prime

$p = 2, {2^2}^{2^{2 - 1}} - 1 = 16 - 1 = 15 \neq$ a prime

$p = 3, {2^2}^{2^{3 - 1}} - 1 = 65536 - 1 = 65535 \neq$ a prime

$p = 4, {2^2}^{2^{4 - 1}} - 1 = 1.15792089×10^{77} - 1 = 1.15792089×10^{77} \neq$ a prime

$p = 5, {2^2}^{2^{5 - 1}} - 1 = 2^{65536} - 1 \neq$ a prime:

Therefore, there is no triple Mersenne prime where $p$ is a prime, however, when $p = 1$ , a non-prime, it produces a prime. I have also discovered Triple Mersenne Numbers!

#NumberTheory

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Comments

@Yajat Shamji, why is it undefined? Its value is still $2 ^ {2 ^ {16}} - 1$

Aryan Sanghi - 1 year ago

Trying to find that out now, I have further simplified it to $2^{65536} - 1$ . @Aryan Sanghi

A Former Brilliant Member - 1 year ago

It is very big, but that is defined, isn't it? Also, maybe at p = 11, it is a prime, but so big we can't evaluate, isn't it?

Aryan Sanghi - 1 year ago

@Aryan Sanghi – What program you got? You sound very confident, @Aryan Sanghi.

A Former Brilliant Member - 1 year ago

@A Former Brilliant Member – Actually above number is defined, codes can't evaluate that much big numbers accurately(I do code and that's why I am saying). Above number is composite for every positive integer p as it can be written of form $a ^ 2 - b ^2$ which can be factored as (a + b)(a - b). I hope I explained well. :)

Aryan Sanghi - 1 year ago

@Aryan Sanghi – Also, I am using Alpertron's Integer Factorization Calculator - best IFC I have ever seen - uses Eliptical Curve Method as well as others.

A Former Brilliant Member - 1 year ago

@A Former Brilliant Member – No calculator in the world can factor it as no code till now written can factor that much big numbers within 1day. 10^7 operations can be performed at max in one second and factoring it requires more than 10^50 operations.

Aryan Sanghi - 1 year ago

@Aryan Sanghi – So, how long? Also, it's not a prime. I'll send the picture.

A Former Brilliant Member - 1 year ago

@A Former Brilliant Member – Yes, I know it is not a prime. As I said for every positive integer p, it can be factored to a number of form (a + b)(a - b). So, actually for p = 4 also, it is not a prime.

Aryan Sanghi - 1 year ago

@Aryan Sanghi – I will check but I believe that $p = 4$ is a prime - let me check first.

A Former Brilliant Member - 1 year ago

@Aryan Sanghi – You're right - I will edit it.

A Former Brilliant Member - 1 year ago

You just checked for numbers 2,3,4,5 then how do you know that there can't be any prime after that?

Vinayak Srivastava - 1 year ago

As it can be written of form a² - b² which can be factored into (a + b)(a - b)

Aryan Sanghi - 1 year ago

That is OK, but the proof in the note is not valid in my opinion.

Vinayak Srivastava - 1 year ago

@Vinayak Srivastava – Yes, actually previously it was that at p= 4 it's prime. After that he changed it to that it is not prime after I told him. So, it's ok

Aryan Sanghi - 1 year ago

Just checking for some numbers is not a proof, I think it should be done algebraically.

Vinayak Srivastava - 1 year ago

You can read the long discussion we had in previous comment.

Aryan Sanghi - 1 year ago

@Yajat Shamji, good job!! I never thought of p = 1.

Aryan Sanghi - 1 year ago

@Aryan Sanghi, what was the proof you were telling of there? I wish to know it!

Vinayak Srivastava - 1 year ago

Simply that 2^2^2^(p-1) can be written as a² and 1 can be written as 1² which can be factored to (a + 1)(a -1).

Aryan Sanghi - 1 year ago

I am making a note on Quadruple Mersenne Numbers and Primes. What do you think, @Aryan Sanghi, @Vinayak Srivastava?

A Former Brilliant Member - 1 year ago

I guess you should rather post a question on it.

Aryan Sanghi - 1 year ago

What should the question be? @Aryan Sanghi

A Former Brilliant Member - 1 year ago

@A Former Brilliant Member – Anything like whether a quadruple prime exists or how many quadruple primes are possible.

Aryan Sanghi - 1 year ago

@Aryan Sanghi – Thanks - was waiting for a reply...

A Former Brilliant Member - 1 year ago

What should the question be? @Vinayak Srivastava

A Former Brilliant Member - 1 year ago

In mathematics it is not sufficient to check only some cases to prove something, either you check for all primes (which are infinite), or either you give a mathematical proof so that the statement can mathematically be accepted.

Zakir Husain - 1 year ago

@Zakir Husain, thank you so much for reminding me but I think you should ask @Aryan Sanghi. He has got the proof.

A Former Brilliant Member - 1 year ago

@Mahdi Raza, @Alak Bhattacharya, @Gandoff Tan, @Hamza Anushath, @Aryan Sanghi

A Former Brilliant Member - 1 year ago

Will nobody comment on the fact that I have found Triple Mersenne numbers? @Vinayak Srivastava, @Aryan Sanghi

A Former Brilliant Member - 1 year ago

Can you tell any such prime you have discovered? :)

Vinayak Srivastava - 1 year ago

I haven't discovered any prime as of yet.

A Former Brilliant Member - 1 year ago

@A Former Brilliant Member – Then????

Vinayak Srivastava - 1 year ago

@Vinayak Srivastava – I am working on it - currently $2^{131072} - 1$ is undefined according to all calculators or is above their amount of numbers limit.

A Former Brilliant Member - 1 year ago

@A Former Brilliant Member – It is not prime. As @Aryan Sanghi said, $2^{131072}$ is a square and so it will be a difference of two squares. Also, $a-b\neq 1$ , so it will have at least 2 factors other than itself and 1.

Vinayak Srivastava - 1 year ago

@Vinayak Srivastava – Can we stop talking about Triple Mersenne primes?! I have discovered Triple Mersenne numbers!

A Former Brilliant Member - 1 year ago

@Vinayak Srivastava – Not 2, but 20 + factors!

A Former Brilliant Member - 1 year ago

@Vinayak Srivastava – There won't be any prime of such type, except when $n=1$ , then $2^{2^{{2^{0}}}}-1=2^{{2^1}}-1=4-1=3$ , which is prime. I don't know of any other case.

Vinayak Srivastava - 1 year ago

@Vinayak Srivastava – LaTeX, please! (I can't see the red bit.)

A Former Brilliant Member - 1 year ago

@A Former Brilliant Member – Haha, I have corrected it, only one extra { messes things!

Vinayak Srivastava - 1 year ago

@Vinayak Srivastava – Thank god! There is a Triple Mersenne Prime I am editing the note right now!

A Former Brilliant Member - 1 year ago

@A Former Brilliant Member – :)

Vinayak Srivastava - 1 year ago

@A Former Brilliant Member – It is impossible. I'll post a reason. Stay tuned.

Aryan Sanghi - 1 year ago

@Aryan Sanghi – But what is the flaw in my comment? (I know it, you know it, I want @Yajat Shamji to see.)

Vinayak Srivastava - 1 year ago

@Aryan Sanghi – I know $1$ is not a prime - look at the note now!

A Former Brilliant Member - 1 year ago

@A Former Brilliant Member – The first sentence itself contradicts it!

$p>1$

Vinayak Srivastava - 1 year ago

@Vinayak Srivastava – Good one! But look at my title...

A Former Brilliant Member - 1 year ago

@A Former Brilliant Member – I saw it, and instantly knew it isn't possible except 1, and 1 isn't prime!

Vinayak Srivastava - 1 year ago

@Vinayak Srivastava – Now, look at the first sentence..

A Former Brilliant Member - 1 year ago

@A Former Brilliant Member – Oh you changed it! :)

Vinayak Srivastava - 1 year ago

@Vinayak Srivastava – Yes - the first sentence actually contradicts the title!

A Former Brilliant Member - 1 year ago

@A Former Brilliant Member – No problem, still a good effort!

Vinayak Srivastava - 1 year ago

@Vinayak Srivastava – Also, can we stop talking about Triple Mersenne primes?! I have discovered Triple Mersenne numbers!

A Former Brilliant Member - 1 year ago

@A Former Brilliant Member – I really appreciate your efforts. As you can see in comments, I was at your side. Keep it up.:)

Aryan Sanghi - 1 year ago

@A Former Brilliant Member – I am no way criticizing you, don't feel that! I am glad and really appreciate your work!

Vinayak Srivastava - 1 year ago

@Vinayak Srivastava – I wasn't talking about criticism.

A Former Brilliant Member - 1 year ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$

Triple Mersenne Primes Proof where \(p\) is either prime or non-prime and On the Matter of Triple Mersenne Numbers

Comments