I don't understand this formula ..(((2^p)-2)/p)^2

Godzilla · 2019-03-12, 08:28

Good morning ,

$(\frac{2^{p}-2}{p})^{2}$

I tried this formula and it seems that if the number that returns the result is an integer it happens that p is a prime number, vice versa if it is a float number p = is not a prime number.

good day to you.

R. Gerbicz · 2019-03-12, 08:48

Try p=341 (Frédéric Sarrus, 1819!).

paulunderwood · 2019-03-12, 08:51

Quote:

Originally Posted by Godzilla

Good morning ,

$(\frac{2^{p}-2}{p})^{2}$

I tried this formula and it seems that if the number that returns the result is an integer it happens that p is a prime number, vice versa if it is a float number p = is not a prime number.

good day to you.

Code:

? ((2^341-2)/341)^2
172563239398117273000218395474248723524247825368023692410729784705364145650902467350956998726199306256798037938929706033772341405068055342449427015484989122022407614607963613122004059079503254688702500

Is 341 prime?

See: https://en.wikipedia.org/wiki/Fermat_pseudoprime

ATH · 2019-03-12, 09:01

http://mathworld.wolfram.com/FermatsLittleTheorem.html

Only for primes and pseudoprimes is p a factor of 2^p-2 which makes (2^p-2)/p an integer.

Godzilla · 2019-03-12, 09:08

Thanks to all of you, now I remember trying a similar formula but giving mod 2, I hadn't noticed the similarity.

axn · 2019-03-12, 09:34

The squaring is useless. It can't turn a fraction into an integer. So, you only need to check if 2^p==2 (mod p). Which is Fermat's little theorem with base = 2

2019-03-12, 08:28	#1
Godzilla May 2016 2·3⁴ Posts	I don't understand this formula ..(((2^p)-2)/p)^2 Good morning , $(\frac{2^{p}-2}{p})^{2}$ I tried this formula and it seems that if the number that returns the result is an integer it happens that p is a prime number, vice versa if it is a float number p = is not a prime number. good day to you. Last fiddled with by Godzilla on 2019-03-12 at 08:29

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2019-03-12, 08:48	#2
R. Gerbicz "Robert Gerbicz" Oct 2005 Hungary 5A2₁₆ Posts	Try p=341 (Frédéric Sarrus, 1819!).

2019-03-12, 09:01	#4
ATH Einyen Dec 2003 Denmark 101111010111₂ Posts	http://mathworld.wolfram.com/FermatsLittleTheorem.html Only for primes and pseudoprimes is p a factor of 2^p-2 which makes (2^p-2)/p an integer.

2019-03-12, 09:08	#5
Godzilla May 2016 2·3⁴ Posts	Thanks to all of you, now I remember trying a similar formula but giving mod 2, I hadn't noticed the similarity.

2019-03-12, 09:34	#6
axn Jun 2003 12F8₁₆ Posts	The squaring is useless. It can't turn a fraction into an integer. So, you only need to check if 2^p==2 (mod p). Which is Fermat's little theorem with base = 2