Thread: Is this a prime number??

Is (3^9 -1)/2 prime??I'll also need to justify my answer using the method along the lines of group theory or mod but have no idea :S thxxxxx

Originally Posted by bryan06

Is (3^9 -1)/2 prime??I'll also need to justify my answer using the method along the lines of group theory or mod but have no idea :S thxxxxx

(3^9-1)/2=[(3^3)3-1^3]/2=(3^3-1)(3^6+3^3+1)/2=[(3^3-1)/2]*(3^6+3^3+1)

Note that both factors are natural numbers. So, (3^9-1)/2 is not prime.

thank you so much for your reply, much appreciated~~~but im so sorry i am kind of lost already wen i try to understand (3^9-1)/2=[(3^3)3-1^3]/2...do you actually mean that or is there a typo??or have i mis-read something?the number involved in the question is (3^9 -1)/2 which is 9841, just in case there's any notation error. would you be very kind and explain briefly your working pleaseee?? thankssssss

Originally Posted by alexmahone

(3^9-1)/2=[(3^3)3-1^3]/2=(3^3-1)(3^6+3^3+1)/2=[(3^3-1)/2]*(3^6+3^3+1)

Note that both factors are natural numbers. So, (3^9-1)/2 is not prime.

Originally Posted by bryan06

Is (3^9 -1)/2 prime??I'll also need to justify my answer using the method along the lines of group theory or mod but have no idea :S thxxxxx

No, since it is less than 10000 you need only check for divisibility by odd primes less than 100 and there are only 24 of those and you can rule out 3 and 5 straight off. Now proving it by some other method is another question, maybe the Lucas–Lehmer test would do (but it still looks more complicated than trial division by the primes less than 100)

CB

Originally Posted by bryan06

thank you so much for your reply, much appreciated~~~but im so sorry i am kind of lost already wen i try to understand (3^9-1)/2=[(3^3)3-1^3]/2...do you actually mean that or is there a typo??or have i mis-read something?the number involved in the question is (3^9 -1)/2 which is 9841, just in case there's any notation error. would you be very kind and explain briefly your working pleaseee?? thankssssss

I merely factorised the given number (9841=13x757).

-------------------------------------------------------------

3^9=(3^3)^3
1=1^3

I then used the "difference of cubes" formula.

ahhhh i see i see thank you soo much

Originally Posted by alexmahone

I merely factorised the given number (9841=13x757).

-------------------------------------------------------------

3^9=(3^3)^3
1=1^3

I then used the "difference of cubes" formula.

but what happens if we're not allowed a calculator (like in my exam)??becoz then we would never know the number is 9841 and therefore cannot use this method. is there another method possible??perhaps using "mod"?? :S

Originally Posted by bryan06

but what happens if we're not allowed a calculator (like in my exam)??becoz then we would never know the number is 9841 and therefore cannot use this method. is there another method possible??perhaps using "mod"?? :S

I didn't use a calculator when I wrote post #2...

so you mean there was no need to know the number was 9841 in the first place??or did u work it out by hand?? cheers

Originally Posted by alexmahone

I didn't use a calculator when I wrote post #2...

Originally Posted by bryan06

so you mean there was no need to know the number was 9841 in the first place??

Exactly.

Originally Posted by bryan06

or did u work it out by hand??

Nope.

i get it now thx

Originally Posted by alexmahone

Exactly.



Nope.