Are you sure? If $x=2^kr$ then we obtain $k+2\geq 2^kr$ for $k>0$ and $1\geq r$ for $k=0$... Too easy (?) Am I missing something?

yes, it is very easy problem, showing by induction on the max power of $2$ dividing $x$, what is the highest power of $2$ dividing $3^x-1$ easy problem...

Myth wrote: Am I missing something? I said easy diophantine equation ...

i dont get the solution??' details for not so good mexikan aspiring olympic please Click to reveal hidden text

by induction $3^{2^n}\cong 1+2^{n+2}\pmod{2^{n+3}}$ so for $m$ odd $3^{m2^n}\cong 1+2^{n+2}\pmod{2^{n+3}}$. This is easy if you have seen other problems like it. Otherwise, not really. Andrei

i heard that this equation can be solved using fields. can anyone show me how to do that?

so... what do you say?

Here is my solution:3ⁿ-1=2ⁿm ||3ⁿ-1||_2=||3-1||_2+||n||_2=1+||n||_2 ->1+||n||_2=n+||m||_2->trivial

what is $\| \cdot \|_2$ in your solution?

That curious symbol $||n||_2$ for the highest (exponent of a) power of $2$ dividing $n$ cannot be rotted out... Where does it come from¿ Any book¿ I#M strongly against it, since it means something else in my opinion! But why not using $\LaTeX$ and using $m,n$ instead of $x,y$¿ Any reason¿

It's simple I don't like using Latex .

Using $\LaTeX$ is a moral You should use it! E.g., the above isn't very good readable. I just don't want to read such things in the knowledge that the writer could have done better by using $\LaTeX$ and hope other people think so, too. But are you posting here for posting only or with the intention that somebody reads and understands it¿ $\LaTeX$ is standard here, and most Mathlinkers use it. There is as good no reason for not using it, especially not something as 'I don't want to learn/use it'. Most agree with the use of $\LaTeX$, it's accepted widely here. So use it please.

We don't like people NOT using LaTeX...

Sorry,guys.My teacher have just teached me how to write 3^n as above and something which is similar to that.So I only want to try write it here..

could some of you guys explain your solutions a little better, i really don't understand, myth, you use the substitution $x=2^k*r$ i suppose the most important question is: What made you think of that substitution? it really seems kind of random to me... also i don't see how this: (putting in the substitution) $3^{2^{k}r}=2^{2^{k}r}y+1$ implies $k+2\ge 2^{k}r$ huh? where did the +2 come from? sorry if these are stupid questions, but i completely do not understand what you did and i want to learn...

Maybe you understand what is written in the following PDF. I can't be sure of what made Myth think of that substitution, but I think it is experience. Many people seem to know that trick. I didn't know it until this problem.

Attachments:
1.pdf (66kb)

perfect_radio wrote: Maybe you understand what is written in the following PDF. I can't be sure of what made Myth think of that substitution, but I think it is experience. Many people seem to know that trick. I didn't know it until this problem. thanks, that was a very good solution, i understand it perfectly, i will still put some thought into the $3^{2^n}-1$ stuff, that seems to be a useful technique, and has a few nice results associated with it

Mind you, that's not my creation. (I wrote that PDF, but I think you understand what I mean) Many people solved it at that TST and used the exact same method. It seems that idea Myth used is unusually well-known.

perfect_radio wrote: Mind you, that's not my creation. (I wrote that PDF, but I think you understand what I mean) Many people solved it at that TST and used the exact same method. It seems that idea Myth used is unusually well-known. i realize, but you took the time to type it up

If x=1 then (x,y)=(1,1) If x>1 then $ 3 \equiv -1 (mod 4)$ and $2^x y +1 \equiv 1 (mod 4) $ so x is even. If x>6 then we have repetition of $ x \not\equiv 0 (mod 2^x) $ Thus we have solutions (x,y)= (2,2), (4,5), (1,1)

if x is odd we get $3^x - 1\equiv 2 (mod 4)$ thus $x\leq1$ and we get $(x,y)=(1,1)$ if x is even , from $LTE$ we have $v_2(3^x-1^x)=v_2(2)+v_2(4)+v_2(x)-1=v_2(x)+2$ thus , $v_2(x)=x-2$ and let $x=2^t*i$ where i is an odd positive integer we get $t=2^t*i-2$ which is impossible for $3\leq i$ , thus $i=1$ we see inductively that $t<2^t-2$ for $3\leq t$ since we were in the case where x is even , we only have to check $t=1$ and $t=2$ both of them work , thus we get the solutions $(x,y)=(2,2)$ and $(x,y)=(4,5)$