A theorem for fun

For discussing Olympiad Level Number Theory problems
User avatar
Abdul Muntakim Rafi
Posts:173
Joined:Tue Mar 29, 2011 10:07 pm
Location:bangladesh,the earth,milkyway,local group.
A theorem for fun

Unread post by Abdul Muntakim Rafi » Tue Sep 06, 2011 12:07 pm

I have recently found this theorem while solving a problem... I was trying to solve the problem... I was thinking about different cases... And I came up with the theorem... :? Can you prove it?


\[(2a+1)^{2b+1}\equiv 2a+1 (mod 2c)\]

and \[(2a+1)^{2b}\equiv 1 (mod 2c)\]
which is the same thing as the above one...

Here $a,b,c$ are non negative integers... The theorem is true for some non negative values of
$a,b,c$ ... :?
I think if we find a way to determine what the values of $a,b,c$ are this might be considered as a theorem... :shock: :shock: :shock: :? :? :? :roll: :roll: :roll:
Man himself is the master of his fate...

User avatar
*Mahi*
Posts:1175
Joined:Wed Dec 29, 2010 12:46 pm
Location:23.786228,90.354974
Contact:

Re: A theorem for fun

Unread post by *Mahi* » Tue Sep 06, 2011 2:50 pm

Abdul Muntakim Rafi wrote: \[(2a+1)^{2b+1}\equiv 2a+1 (mod 2c)\]

and \[(2a+1)^{2b}\equiv 1 (mod 2c)\]
You say something like this is a theorem? Like solving some congruence like $x^y \equiv 1 (mod z)$ where $x$ is odd and $y,z$ are even? It might be a problem, but a theorem consists of much more things. And there is much in fact, to solve conguences like $x^y \equiv 1 (mod z)$.
Please read Forum Guide and Rules before you post.

Use $L^AT_EX$, It makes our work a lot easier!

Nur Muhammad Shafiullah | Mahi

User avatar
afif mansib ch
Posts:85
Joined:Fri Aug 05, 2011 8:16 pm
Location:dhaka cantonment

Re: A theorem for fun

Unread post by afif mansib ch » Wed Sep 07, 2011 12:41 am

am i getting the thing wrong or it is easy?i think i'm wrong with something :mrgreen:

we know for any constant a and c
\[(2a)\equiv0(mod2c)\]
and (2a+1)^(2b)is congruent to 1(mod 2c)
now multiplying both sides by (2a+1) we get
(2a+1)^(2b+1)is congruent to 2a+1(mod 2c)
is it all?
[i am really sorry to write down like this.i am not being able to learn using latex properly.so i wrote like this :mrgreen: ]

User avatar
Abdul Muntakim Rafi
Posts:173
Joined:Tue Mar 29, 2011 10:07 pm
Location:bangladesh,the earth,milkyway,local group.

Re: A theorem for fun

Unread post by Abdul Muntakim Rafi » Wed Sep 07, 2011 11:37 am

Well that's why I called it a theorem for fun... You can see it as a problem...

And Afif, first you have to prove that
(2a+1)^(2b) is congruent to 1(mod 2c)
then we are done...
By the way,I used this theorem(!) while solving your congruence problem... :D
Man himself is the master of his fate...

User avatar
*Mahi*
Posts:1175
Joined:Wed Dec 29, 2010 12:46 pm
Location:23.786228,90.354974
Contact:

Re: A theorem for fun

Unread post by *Mahi* » Wed Sep 07, 2011 2:05 pm

afif mansib ch wrote: we know for any constant a and c
\[(2a)\equiv0(mod2c)\]
Is it? What about $a=3,c=4$ and infinite more $a,c$
Please read Forum Guide and Rules before you post.

Use $L^AT_EX$, It makes our work a lot easier!

Nur Muhammad Shafiullah | Mahi

User avatar
afif mansib ch
Posts:85
Joined:Fri Aug 05, 2011 8:16 pm
Location:dhaka cantonment

Re: A theorem for fun

Unread post by afif mansib ch » Wed Sep 07, 2011 4:03 pm

sorry mahi via i didn't notice it. :? watching 2 in both sides i wrote that. :D then what should i write? :mrgreen:

User avatar
afif mansib ch
Posts:85
Joined:Fri Aug 05, 2011 8:16 pm
Location:dhaka cantonment

Re: A theorem for fun

Unread post by afif mansib ch » Thu Sep 08, 2011 12:38 pm

i wanted to say if one of the statement is true by multiplying or dividing it by (2a+1)we can get the other one.so what else is needed to be proved? :?

User avatar
*Mahi*
Posts:1175
Joined:Wed Dec 29, 2010 12:46 pm
Location:23.786228,90.354974
Contact:

Re: A theorem for fun

Unread post by *Mahi* » Thu Sep 08, 2011 5:28 pm

Okay you might need to know some of the basics about congruence, so please see this post, and try some textbooks.
Please read Forum Guide and Rules before you post.

Use $L^AT_EX$, It makes our work a lot easier!

Nur Muhammad Shafiullah | Mahi

User avatar
afif mansib ch
Posts:85
Joined:Fri Aug 05, 2011 8:16 pm
Location:dhaka cantonment

Re: A theorem for fun

Unread post by afif mansib ch » Thu Sep 08, 2011 10:28 pm

i am act reading the book ELEMENTARY NUMBER THEORY.thanks for the suggestion.have i got the ques wrong?i think he meant to proof if one of the statement is true then the other one is also.so do i need to proof anything else? :cry:

User avatar
*Mahi*
Posts:1175
Joined:Wed Dec 29, 2010 12:46 pm
Location:23.786228,90.354974
Contact:

Re: A theorem for fun

Unread post by *Mahi* » Thu Sep 08, 2011 10:54 pm

I think you do have the question wrong...it's about just some cases of when those two things would happen.
BTW you may also try Number Theory by S.G.Telang as a text, it is also good for beginning problem solvers like you.
Please read Forum Guide and Rules before you post.

Use $L^AT_EX$, It makes our work a lot easier!

Nur Muhammad Shafiullah | Mahi

Post Reply