Matematické Fórum

stuart clark · 06. 12. 2011 16:03

the number $\frac{2^{58}+1}{5}$ is prime no. or Composite no.

I want proof.

thank

Pavel · 06. 12. 2011 19:39 — Editoval Pavel (08. 12. 2011 20:20)

↑ stuart clark:

It is a wrong solution.

stuart clark · 07. 12. 2011 17:49

thanks pavel got it

i have used sophy germain Identity

Marian · 08. 12. 2011 17:40

↑ Pavel:

How can we see that the considered fraction is not prime. Your proof shows (only?) that the fraction is a positive integer.

vanok · 08. 12. 2011 18:01 — Editoval vanok (09. 12. 2011 10:07)

Hi ↑ Marian:,
The formula of Sophie Germain is: $x^4 + 4y^4=((x+y)^2 +y^2) ((x-y)^2 +y^2)$

Autor skryl část textu. Zobrazit/skrýt!

What gives us with Paul's observation one factorisation.
Sincerely Vanok.

Pavel · 08. 12. 2011 20:19 — Editoval Pavel (08. 12. 2011 20:19)

↑ Marian:

Yes, you are right. It is my mistake, I did not realize that that is the proof of the fact that the fraction is only a positive integer. I add a note to the solution.

stuart clark · 09. 12. 2011 06:35

Autor skryl část textu. Zobrazit/skrýt!

stuart clark · 09. 12. 2011 06:41

↑ vanok:

would you like to explain me why the given number is composite

$\frac{2^{58}+1}{5}=536903681.107367629$

thanking you

vanok · 09. 12. 2011 10:36

Hi ↑ stuart clark:,
I calculate simply (feasible simple calculation manually):
$2^{14}$
$2^{28}=(2^{14})^2$
$(2^{14}+1)^2$
$(2^{14}+1)^2+2^{28}=536903681$
$(2^{14}-1)^2$
$(2^{14}-1)^2+2^{28}=536838145$
and
$\frac {(2^{14}-1)^2+2^{28}}5=107367629$
Then
$1^4 +4. (2^{14})^4=((2^{14}+1)^2+2^{28})((2^{14}-1)^2+2^{28})=536903681.536838145$
and finally
$\frac{2^{58}+1}{5}= (2^{14}+1)^2+2^{28})(\frac {(2^{14}-1)^2+2^{28}}5)=536903681.107367629$
Sincerely Vanok