## ONTC Final Exam

Discussion on Bangladesh National Math Camp
Epshita32
Posts: 37
Joined: Mon Aug 24, 2015 12:34 am

### Re: ONTC Final Exam

In p5 can I use the cyclic version instead of the given one ?

Epshita32
Posts: 37
Joined: Mon Aug 24, 2015 12:34 am

### Re: ONTC Final Exam

In p2 , 2^m-1 according to m makes a sequence - if m = 1 , then 2^m-1 = 1 . For m as 2 , 2^m-1 = 3 . For m as 3 , 2^m-1 = 7 . Let 2^m-1 = X . The sequence - Xn = 2*X(n-1) + 1 where X1 = 1 . In the whole sequence only when m = 1 , X divides n^2+1 . For ex : n^2+1 = 1, 2 (mod 3) etc . My proof is poor . I can't post it . Can anyone give me any hints ?

Masum
Posts: 592
Joined: Tue Dec 07, 2010 1:12 pm

### Re: ONTC Final Exam

Epshita32 wrote:In p5 can I use the cyclic version instead of the given one ?
Yeah. no problem
One one thing is neutral in the universe, that is $0$.

Masum
Posts: 592
Joined: Tue Dec 07, 2010 1:12 pm

### Re: ONTC Final Exam

Epshita32 wrote:In p2 , 2^m-1 according to m makes a sequence - if m = 1 , then 2^m-1 = 1 . For m as 2 , 2^m-1 = 3 . For m as 3 , 2^m-1 = 7 . Let 2^m-1 = X . The sequence - Xn = 2*X(n-1) + 1 where X1 = 1 . In the whole sequence only when m = 1 , X divides n^2+1 . For ex : n^2+1 = 1, 2 (mod 3) etc . My proof is poor . I can't post it . Can anyone give me any hints ?
Epshita, try to learn latex gradually
Now, as for your proof, it's not really a proof, more like an observation.
Killer hint: A prime divisor of $a^2+b^2$ where $\gcd(a,b)=1$ is of the form $4k+1$. Try now!
One one thing is neutral in the universe, that is $0$.

Epshita32
Posts: 37
Joined: Mon Aug 24, 2015 12:34 am

### Re: ONTC Final Exam

Masum bai , I know this is not a proof . I was talking about my proof that I haven't posted yet . And thank you . My solution consists of the hint you gave . I was thinking of this was correct or not .

Masum
Posts: 592
Joined: Tue Dec 07, 2010 1:12 pm

### Re: ONTC Final Exam

You can post the full solution if you want.
And bai, bai? Seriously?
One one thing is neutral in the universe, that is $0$.

Epshita32
Posts: 37
Joined: Mon Aug 24, 2015 12:34 am

### Re: ONTC Final Exam

When will you post the solutions ? And I was thinking of new ways to solve p2 . Can it be done by congruence ?

Masum
Posts: 592
Joined: Tue Dec 07, 2010 1:12 pm

### Re: ONTC Final Exam

I posted the solutions of the first one in my new blog. But my laziness is getting the better of me again. When that passes, I will start writing again there. Stay tuned lol
One one thing is neutral in the universe, that is $0$.

tanmoy
Posts: 305
Joined: Fri Oct 18, 2013 11:56 pm

### Re: ONTC Final Exam

$\text {Solution of problem 2}$:
Suppose,$p$ is an odd divisor of $n^{2}+1$.So, $p$ is either of the form $4x+1$ or $4x+3$,where $x$ any integer.Let $p$ is of the form $4x+3$.Then $n^{2} \equiv -1 (mod p)$.Or $n^{p-1} \equiv (n^{2})^{2x+1} \equiv -1 (mod p)$,which contradicts Fermat's little theorem.
So,$p$ is of the form $4x+1$.So, $2^{m}-1$ is of the form $4x+1$.So, $m=1$ is the only solution.
"Questions we can't answer are far better than answers we can't question"

Masum
Posts: 592
Joined: Tue Dec 07, 2010 1:12 pm

### Re: ONTC Final Exam

Ya, this is the way to do it.
One one thing is neutral in the universe, that is $0$.