new problem-2 (BOMC-2)

[sm.joty]

Find all $n$ such that $2^n|3^n-1$

[*Mahi*]

Hint:

$9\equiv 1 \text{ (mod }8\text)$+Direct LTE

[Nadim Ul Abrar]

Case 1 : $n$ is odd
Case 2 : $n$ is even

Case 1
$3^2=1mod 4$
$3^n-1 \equiv 3-1 \equiv 2 (mod4)$

So we just have $n=1$ for this case .

Case 2
let $n=2^km$
Now $n \leq v_2(2)+v_2(4)+v_2(n)-1=2+k$
or $k+2 \geq 2^k.m$

this leads possible values of k are , $k=2,1$ and $ m=1$
Now just check . $k=1,2$ both satisfy the condition .

So all solutions are $n=1,2,4$ ;

[Niloy Da Fermat]

Case 1 : $n$ is odd
Case 2 : $n$ is even

Case 1
$3^2=1mod 4$
$3^n-1 \equiv 3-1 \equiv 2 (mod4)$

So we just have $n=1$ for this case .

Case 2
let $n=2^km$
Now $n \leq v_2(2)+v_2(4)+v_2(n)-1=2+k$
or $k+2 \geq 2^k.m$

this leads possible values of k are , $k=2,1$ and $ m=1$
Now just check . $k=1,2$ both satisfy the condition .

So all solutions are $n=1,2,4$ ;

এখানে $ v_2(a) $ দিয়ে কি বোঝানো হয়েছে? ব্যাখ্যা করলে ভাল হয় ।

[Tahmid Hasan]

$u_p(x)=a$ means $p^a \mid x$ but $p^{a+1}$ doesn't divide $x$.
and @Nadim vaiya,please state that yo used LTE in your solution.

[Niloy Da Fermat]

thanks @Tahmid

[Phlembac Adib Hasan]

@ Nadim vaia, I also solved like you.And yes, LTE kills the problem.

[itsnafi]

Can anyone inform me, what is LTE?

[sourav das]

Can anyone inform me, what is LTE?

Forget about LTE, First try it without using any theorem (actually this one doesn't need so much high level tool)

But if you wish to study about it then:
http://www.artofproblemsolving.com/Reso ... rs/LTE.pdf

[itsnafi]

Thnx to sourav da.