# Thread: Demonstration using congruency modulo m

1. ## Demonstration using congruency modulo m

Well, i am stuck with a problem. I have to prove that 7|(3^(2n+1) + 2^(n+2)) using the modulo m congruency.

Well this will mean that 3^(2n+1) + 2(n+2) mod 7 = 0, but i cant find a whay to prove the initial expression.

Any ideeas?

Thanks

2. ## Re: Demonstration using congruency modulo m

$\displaystyle 3^{2n+1}=3\cdot9^n$, and $\displaystyle 9\equiv2\pmod{7}$...

3. ## Re: Demonstration using congruency modulo m

Thanks, i will give it a try.
I will post a little later what i have succeded to solve. And if you will have some free time i will ask you to look at it.
Thanks again.

4. ## Re: Demonstration using congruency modulo m

This should be : (3*9^n +2^n*4 ) eq (3*2^n + 2^n*4) eq 7*2^n mod 7 = 0 -> 7 | (3 ^(2n+1) + 2^(n+2)).

It it is right?

Also any iddeas for prove that 641|(2^32 + 1)?

Thanks

5. ## Re: Demonstration using congruency modulo m

Originally Posted by Adrian111
This should be : (3*9^n +2^n*4 ) eq (3*2^n + 2^n*4) eq 7*2^n mod 7 = 0 -> 7 | (3 ^(2n+1) + 2^(n+2)).

It it is right?
Yes.

Originally Posted by Adrian111
Also any iddeas for prove that 641|(2^32 + 1)?
2^32 is not such a big number... You can also find 2^16 mod 641, then square it and take the remainder again. Or find 1024 mod 641, raise it to power 3 and multiply by 4.