Question

Find all odd primes which \(\Large \frac{2^{p-1}-1}{p} \) would be a perfect square number.

Answer 1

not a very easy question :P

Answer 2

okk
so
\(2^{p-1}-1 =0 \mod p\)
we need to find p that satisfy
\(2^{p-1}-1=p*n^2\)

Answer 3

i had a feeling of there is no p that makes it true ( still my conjecture not sure :P )

Answer 4

no there are some numbers ;)

Answer 5

awww cute , then lemme work a bit

Answer 6

aww 7 work

Answer 7

yes

Answer 8

and 3

Answer 9

brb

Answer 10

Deja Vu! I'll let @Marki figure this out :)

Answer 11

since when u learned french :O

Answer 12

ok
n^2=1 mod 8 since its odd perfect square

Answer 13

\(2^{p-1}-1=p*n^2=p\mod 8\)
so we would have
\(2^{p-1}=p+1\mod 8\)
idk if this work

Answer 14

oh since its odd i should think of 1 mod 2

Answer 15

hmm got it ! so only 3,7 nice

Answer 16

not a nad start! ;)
but prove that only 3,7 would work...

Answer 17

yeah , the only primes i guess there is no more

Answer 18

i'll show u some other time not feeling like doing NT :3

Answer 19

\(\)

Answer 20

haha @mathmath333 style :P

Answer 21

just the bookmark

Answer 22

hmm,a little help:
\(\Large \frac{2^{p-1}-1}{p} = \frac{ (2^{\frac{ p-1 }{ 2 }}-1)(2^{\frac{ p-1 }{ 2 }}+1) }{ p }\)

Answer 23

*

Answer 24

I remember back when I used to be able to do harder problems than most other user on OS. Those days are (thankfully) long gone, as threads like these remind me :)
*bookmark

Answer 25

users*