$M_{M_{M_{p}}} = {2^2}^{2^{p - 1}} - 1, p \geq 1$ 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!

No vote yet

1 vote

Easy Math Editor

This discussion board is a place to discuss our Daily Challenges and the math and science related to those challenges. Explanations are more than just a solution — they should explain the steps and thinking strategies that you used to obtain the solution. Comments should further the discussion of math and science.

When posting on Brilliant:

`*italics*`

or`_italics_`

italics`**bold**`

or`__bold__`

boldNote: you must add a full line of space before and after lists for them to show up correctlyparagraph 1

paragraph 2

`[example link](https://brilliant.org)`

`> This is a quote`

Remember to wrap math in`\(`

...`\)`

or`\[`

...`\]`

to ensure proper formatting.`2 \times 3`

`2^{34}`

`a_{i-1}`

`\frac{2}{3}`

`\sqrt{2}`

`\sum_{i=1}^3`

`\sin \theta`

`\boxed{123}`

## Comments

Sort by:

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

Log in to reply

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

Log in to reply

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?

Log in to reply

@Aryan Sanghi.

What program you got? You sound very confident,Log in to reply

$a ^ 2 - b ^2$ which can be factored as (a + b)(a - b). I hope I explained well. :)

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 formLog in to reply

Log in to reply

Log in to reply

Log in to reply

Log in to reply

$p = 4$ is a prime - let me check first.

I will check but I believe thatLog in to reply

Log in to reply

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

Log in to reply

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

Log in to reply

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

Log in to reply

Log in to reply

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

Log in to reply

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

Log in to reply

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

Log in to reply

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

Log in to reply

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).

Log in to reply

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.

Log in to reply

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

Log in to reply

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

Log in to reply

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

Log in to reply

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

Log in to reply

I haven't discovered any prime as of yet.

Log in to reply

Log in to reply

$2^{131072} - 1$ is undefined according to all calculators or is above their amount of numbers limit.

I am working on it - currentlyLog in to reply

@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

It is not prime. Aswillhave at least 2 factors other than itself and 1.Log in to reply

stop talking about Triple Mersenne primes?! I have discoveredTriple Mersenne numbers!Log in to reply

20 +factors!Log in to reply

$n=1$, then $2^{2^{{2^{0}}}}-1=2^{{2^1}}-1=4-1=3$, which is prime. I don't know of any other case.

There won't be any prime of such type, except whenLog in to reply

Log in to reply

Log in to reply

is a Triple Mersenne PrimeI am editing the note right now!Log in to reply

Log in to reply

Log in to reply

@Yajat Shamji to see.)

But what is the flaw in my comment? (I know it, you know it, I wantLog in to reply

$1$ is not a prime - look at the note now!

I knowLog in to reply

Log in to reply

Log in to reply

Log in to reply

Log in to reply

Log in to reply

Log in to reply

Log in to reply

stop talking about Triple Mersenne primes?! I have discoveredTriple Mersenne numbers!Log in to reply

Log in to reply

Log in to reply

Log in to reply

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

Log in to reply

I guess you should rather post a question on it.

Log in to reply

What should the question be? @Aryan Sanghi

Log in to reply

Log in to reply

Log in to reply

What should the question be? @Vinayak Srivastava

Log in to reply