Question

What's the idea to solve for m the equation m(3^(2m)+3) = 0 mod 7 ?

Answer 1

3antarrr kefak

Answer 2

so2alak heek ?
\(\large m(3^{2m}+3) = 0 \mod 7 \)

Answer 3

Ahlan ikram, Exactly

Answer 4

what i can see is since 7 is prime humm
let me think :O

Answer 5

ok 3antar 7 is prime 3ashan heek e7na 3ena two cases :-
m=0 mod 7
or
3^(2m)+3 = 0 mod 7

Answer 6

case 1
m=7k

Answer 7

Yes

Answer 8

Why the two cases @ikram002p ?

Answer 9

case 2
3^(2m)+3 = 0 mod 7
we need to show which m makes i true

since 3 is prime then
3^2m mod 7=3 mod 7
so
3 mod 7 + 3 = 6 mod 7 \(\neq\) 0 mod 7
so the only case is m=2k

Answer 10

7 is prime , u cant write it as a multiple of two numbers

Answer 11

Ah' ok

Answer 12

@ikram002p What happened when 3 was prime ?

Answer 13

@rational have you an idea ?

Answer 14

case1 is trivial :
m = 7k

Answer 15

case2 :

You need to solve \(3^{2m}+3 \equiv 0 \mod 7\)

Answer 16

you familiar with Fermat little thm ?

Answer 17

\(3^{2m}+3 \equiv 0 \mod 7\)

\(\implies (3^2)^{m} \equiv 4 \mod 7\)

\(\implies 2^{m} \equiv 4 \mod 7\)

\(\implies 2^{m-2} \equiv 1 \mod 7\)

Answer 18

see if that looks okay so far

Answer 19

@rational Awesome go ahead

Answer 20

are you going to use log base 2, :) ?

Answer 21

haha much simpler than that

Answer 22

one sec

Answer 23

\(\implies 2^{m-2} \equiv 1 \mod 7\)

By FLT, \(m-2 \equiv 0 \mod 6 \implies m = 6k+2\)

gives u solutions, but these are not exhaustive. there will be other solutions as well.

Answer 24

\(m = 7k, ~ 6k+2\) are part of solutions.
let me think a bit to conclude on other solutions

Answer 25

wil get back in the morning, this wil take time :)

Answer 26

@rational How did you use the FLT ?

Answer 27

\(\large a^{p-1} = 1 \mod p\)

Answer 28

IF \(m-2 = 6\),
then it will be a solution of : \(2^{m-2} = 1 \mod 7\), right ?

Answer 29

What's p in our case ? 7, right ?

Answer 30

= or equiv ?

Answer 31

equiv

Answer 32

and yes p=7