Small math question

Question

Small math question

anonymous · Answer

im big so i got this

dangerousjesse · Answer

Can a prime $p$ satisfy \(2^{p − 1} ≡ 1 (mod p^2) and 3^{p − }1 ≡&nbps;1 (mod p^2) simultaneously?

ganeshie8 · Answer

$$\large 2^{p − 1}  \equiv 1 \mod p^2 $$
$$\large 3^{p − 1}  \equiv 1 \mod p^2 $$

ganeshie8 · Answer

like that ?

dangerousjesse · Answer

Yes :). It's just a personal hypothetical question.

ikram002p · Answer

so you know for p alone it can ?

dangerousjesse · Answer

Yes

ganeshie8 · Answer

she wants the same prime number to satisfy both equations

ikram002p · Answer

oh so her question what p that satisfy ?

dangerousjesse · Answer

More or less asking if there's such a thing as a prime number that can satisfy both equations @ikram002p

ikram002p · Answer

why not :)
easy bezy , chineezy :P

ganeshie8 · Answer

chinese remainder theorem ?

ikram002p · Answer

when both equations satisfies then we only wanna find p such that

6^p-1 =1 mod p

it aint gonna work in general since 6 is not perfect square , so u have two methods ,
1_guessing :)
2_chinese remainder theorem

ikram002p · Answer

p^2 *

dangerousjesse · Answer

Yes haha @ganeshie8  "On Lerch's formula for the Fermat quotient"

ikram002p · Answer

ok sound no solution :o
hehe :P

ganeshie8 · Answer

looks hard >.<

ganeshie8 · Answer

It seems there are no solutions the first equation itself :

$$\large 2^{p − 1}  \equiv 1 \mod p^2$$

ikram002p · Answer

well yes :)

ganeshie8 · Answer

I am not able to find any $p$ satisfying above equation itself

ganeshie8 · Answer

can we prove the first equation itself has no solution ?

ikram002p · Answer

wait no

ikram002p · Answer

p=3