Mangammal primes
 

MANGAMMAL PRIMES

To know more about these Google OEIS And see SEQ A123239

A.K.Devaraj

Mangammal primes": prime numbers which are impossible factors of 3^n-2, i.e. they do not divide 3^n-2 for any value of n.

2, 3, 11, 13, 37, 41, 59, 61, 67, 73, 83, 103, 107, 109, 131, 151, 157, 179, 181, 193, 227, 229, 251, 271, 277, 307, 313, 347, 349, 367, 373, 397, 419, 421, 433, 443, 467, 491, 523, 541, 547, 563, 577, 587, 613, 619, 659, 661, 673, 683, 709, 733, 757, 761

COMMENT

That the sequence is infinite can be proved using a theorem in "Euler's Generalisation Of Fermat's Theorem - A Further Generalisation" (ISBN 1550-3747). A.K.Devaraj.

Is there a mersenne prime other than 3 that is also a mangammal prime?

[QUOTE=Damian;88882]Mangammal primes": prime numbers which are impossible factors of 3^n-2, i.e. they do not divide 3^n-2 for any value of n.

[...]

Is there a mersenne prime other than 3 that is also a mangammal prime?[/QUOTE]
Not for the first Mersenne primes:

Let p be the Mersenne exponent.
Then 3^n = 2 (mod 2^p-1) for:

p=3 -> n=1*(2^3-2)/3
p=5 -> n=4*(2^5-2)/5
p=7 -> n=4*(2^7-2)/7
p=13 -> n=12*(2^13-2)/13
p=17 -> n=8*(2^17-2)/17
p=19 -> n=12*(2^19-2)/19
p=31 -> n=28*(2^31-2)/31
p=61 -> n=11*(2^61-2)/61
p=89 -> n=47*(2^89-2)/89
p=107 -> n=99*(2^107-2)/107
p=127 -> n=24*(2^127-2)/127
p=521 -> n=363*(2^521-2)/521
p=607 -> n=597*(2^607-2)/607
p=1279 -> n=600*(2^1279-2)/1279
p=2203 -> n=1026*(2^2203-2)/2203
p=2281 -> n=241*(2^2281-2)/2281
p=3217 -> n=1894*(2^3217-2)/3217
p=4253 -> n=837*(2^4253-2)/4253

I could not continue because UBASIC overflowed.

All congruences below are modulo [tex]2^p-1[/tex]. Also [tex]2^p-1 \,>\, 3[/tex].

First of all, we will prove that there are p different numbers which raised to the p power are congruent to 1:

Let's consider the numbers [tex]2^s[/tex] for 0<s<p. These p numbers are all different because they are less than [tex]2^p-1[/tex].

[tex](2^s)^p \,\eq\, (2^p)^s\, \eq\, 1^s \,\eq \,1[/tex] (1)

So these are p p-roots of 1. Since there cannot be more than p p-roots of 1, all p-roots of 1 must be powers of 2.

Since 3 is not a power of 2, [tex]3^p \not\equiv 1[/tex].

From Fermat's theorem: [tex]3^{(2^p-2)}\,\eq\,1[/tex], this implies [tex]3^{(2^p-2)/p} \eq u \,\not\equiv\, 1 [/tex]

Since [tex]u^p \eq 1[/tex] we find from (1) that [tex]u=2^s[/tex].

Since [tex]\gcd(s, p) = 1[/tex], we get [tex]u^{(1/s)} \,\eq\, 2[/tex].
(where 1/s is the inverse multiplicative of s modulo p).

This shows that a Mersenne prime greater than 3 cannot be a Mangammal prime.

[QUOTE=alpertron;88956]Let's consider the numbers [tex]2^s[/tex] for 0<s<p. These p numbers are all different because they are less than [tex]2^p-1[/tex].[/QUOTE]

It must be [tex]2^s[/tex] for [tex]0\,\le\,s\,<\,p[/tex].

or how about this? :)

2^p-1 prime
=> 3^(2^p-1) == 3 [mod 2^p-1] [Fermat]
=> 3^(2^p-1) - 2 == 1 [mod 2^p-1]
=> gcd (2^p-1, 3^(2^p-1) - 2) = 1

QED,
J

[QUOTE=bearnol;88967]or how about this? :)

2^p-1 prime
=> 3^(2^p-1) == 3 [mod 2^p-1] [Fermat]
=> 3^(2^p-1) - 2 == 1 [mod 2^p-1]
=> gcd (2^p-1, 3^(2^p-1) - 2) = 1

QED,
J[/QUOTE]

Actually, I'm not sure that's complete (I think I have misunderstood the original problem :( ) - perhaps someone can complete this, my line of reasoning, for me???
J

[QUOTE=bearnol;88967]or how about this? :)

2^p-1 prime
=> 3^(2^p-1) == 3 [mod 2^p-1] [Fermat]
=> 3^(2^p-1) - 2 == 1 [mod 2^p-1]
=> gcd (2^p-1, 3^(2^p-1) - 2) = 1

QED,
J[/QUOTE]

I don't understand where you show that there exist an exponent e such that

[tex]3^e\,\eq\,2\,\pmod{2^p-1}[/tex]

[QUOTE=alpertron;88969]I don't understand where you show that there exist an exponent e such that

[tex]3^e\,\eq\,2\,\pmod{2^p-1}[/tex][/QUOTE]

no, i don't think I do :)

Just as an aside - do we know for sure that this result is true - ie does there definitely exist a (longer?) proof. Are you moreorless happy with yours, Dario, for instance? [if so, I'll shut up!]

[QUOTE=bearnol;88971]no, i don't think I do :)

Just as an aside - do we know for sure that this result is true - ie does there definitely exist a (longer?) proof. Are you moreorless happy with yours, Dario, for instance? [if so, I'll shut up!][/QUOTE]

I think my proof is OK. Let's wait until a professional mathematician comes to this thread and says if it is right.

This is the step I don't understand, why is [tex] u \,\not\equiv\, 1 [/tex] ?

[QUOTE=alpertron;88956][tex]3^{(2^p-2)/p} \eq u \,\not\equiv\, 1 [/tex]
[/QUOTE]