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RMO 2015

I have taken inspiration from my friend Swapnil's note and have decided to post this note.I have decided that I will post one or two problems every now and then that are related to the topics of RMO.They can either be proof problems, like the proof of an integral theorem e.t.c or ones like finding values.The questions will be either from preparation books or from Mathematical olympiads.I will keep on adding them one-by-one in the space below.The main rule is that there should be just one solution to one problem,unless,of course,there are more than one way of doing it.If the solution you have is the same as the one which has already been posted,kindly refrain from posting it,but if you have another method of solving the same problem,please do post it!I will post the solution only if one hasn't been posted in \(3\) days.So,\[\text{Happy problem solving!}\]\[\] 1.If p is a prime number,then prove that,\((a+b)^p\equiv(a^p+b^p)\pmod{p}\) Generalize it too!(Awesome part!)\[\] 2.For \(m>2\),prove that,\(\phi(m)\)is even where\(\phi\) is the Euler's Totient Function\[\] 3.Prove that there are infinitely many squares in the sequence \(1,3,6,10,15,21,28,......\).\[\] 4.Find all the pairs of positive integers (m,n) such that \(2^m+3^n\) is a perfect square.\(\text{INMO}\)\[\] 5.Prove that \(2^p+3^p\) is not a perfect square for a prime \(p\).\[\] 6.If \(a\equiv b\pmod{m^n}\),then prove that \(a^m\equiv b^m\pmod{m^{n+1}}\)(Given by Svatejas Shivakumar). \[\] 7.Find a polynomial with integer co-efficients such that \(P(a)=b,P(b)=c,P(c)=a\) where \(a,b,c\) are distinct.(given by Anik Mandal).We need a solution for this.Surya Prakash has posted a solution.\[\] 8.Find the number of prime \(n\) satisfying the equation \(3n+1=k^2,k\in \mathbb{Z}^+\).(Given by Mehul Arora).\[\] 9.If \(x^3=x+1\) then find integers \(a,b,c\) such that \(x^7=ax^2+bx+c\).(Given by Dev Sharma).

Note by Adarsh Kumar
2 years ago

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5) We consider case when \(p=2\). Clearly \(2^2+3^2 = 4+9 = 13\) which is not a perfect square.So now consider \(p>2\). Let if possible \(2^p+3^q=q^2\) for some integer \(q\).We note that \(2^p \equiv 0 \pmod{4}\) , \(3^p \equiv 3 \pmod{4}\) , thus \(2^p+3^p \equiv 3 \pmod{4}\) but \(q^2 \equiv 0,1 \pmod{4}\) which is a contradiction.

Nihar Mahajan - 2 years ago

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Thanx for your solution.

Adarsh Kumar - 2 years ago

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9)\(x^3=x+1\Rightarrow x^6=(x^2+2x+1)\Rightarrow x^7=x^3+2x^2+x=2x^2+2x+1\).

Adarsh Kumar - 2 years ago

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How u got this......explain

Ayush Verma - 1 year, 9 months ago

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9) \(x^3=x+1 \Rightarrow x^7=x^4(x+1) = x^5+x^4 = x^2(x+1)+x(x+1) = x^3+x^2+x^2+x = 2x^2+2x+1 \\ \Rightarrow (a,b,c)=(2,2,1)\)

Nihar Mahajan - 2 years ago

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No need of all that.See my solution.

Adarsh Kumar - 2 years ago

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7) If \(P(x)\) is a polynomial with integer coefficients then for distinct integers \(a\), \(b\), we have \(a-b|P(a)-P(b)\).

So, it implies that \(a-b|P(a)-P(b) = b-c\). Similarly we get, \(b-c|c-a\) and \(c-a|a-b\). So finally what we get is \(a-b|b-c\), \(b-c|c-a\) and \(c-a | a-b\), this is possible iff \(a-b = b-c =c-a\). But this implies that \(a=b=c\). This is a contradiction as \(a\), \(b\) and \(c\) are distinct. Therefore no such polynomial exists.

Surya Prakash - 2 years ago

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Is the first statement of your solution a lemma?

Nihar Mahajan - 2 years ago

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Yes. It is a lemma. it is easy to prove

Surya Prakash - 2 years ago

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@Surya Prakash Can you give me hint to prove it? Is it necessary to define a degree to the polynomials?

Nihar Mahajan - 2 years ago

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@Nihar Mahajan Take \(P(x) = a_{n} x^n + a_{n-1} x^{n-1} + \ldots + a_{0}\) where \(a_{n} , a_{n-1} , \ldots a_{0}\) are integers. Now, subtract \(P(a)\) from \(P(b)\) and observe what happens.

Surya Prakash - 2 years ago

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No,it can be proved like this,\[Since\ a-b|b-c,a-b\leq b-c\\ Since\ b-c|c-a,b-c\leq c-a\\ Since\ c-a|a-b,c-a\leq a-b\].Combining these you get that the equality holds.

Adarsh Kumar - 2 years ago

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Thanx for the solution.

Adarsh Kumar - 2 years ago

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Thanks for the solution!

Anik Mandal - 2 years ago

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6) \(a^{m}-b^{m}=(a-b)(a^{m-1}+a^{m-2}b+\ldots+ab^{m-2}+b^{m-1})\).

Since \(a \equiv b \pmod {m^{n}},a \equiv b \pmod {m}\).

Hence, \(a^{m-i}b^{i} \equiv b^{m} \pmod {m}\) and \(a^{m-1}+a^{m-2}b+\ldots+ab^{m-2}+b^{m-1} \equiv mb^{m} \equiv 0 \pmod {m}\)

Hence, \(a^{m}-b^{m}\) is divisible by \(m^{n+1}\).

Note: This solution is not original.

Credit: An Excursion in Mathematics.

Brilliant Member - 1 year, 12 months ago

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Really elegant one! Thanx for posting it!

Adarsh Kumar - 1 year, 12 months ago

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Your welcome :)

Brilliant Member - 1 year, 12 months ago

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For the \(3rd\) question,

The sequence is, \(S=1+3+6+10+15........+{t}_{n} \)

To get the \( {t}_{n} \) we can write it as,

\(S=0+1+3+6+10+15........+{t}_{n} \)

\(S=1+3+6+10+15+.....\)

Now, Subtracting both the sums,

\(0=-1-2-3-4-5-.......{t}_{n} \)

Therefore we can get,

\( {t}_{n}=\frac {n(n+1)}{2} \)

Now to prove that there are infinitely many squares, Assume that \({t}_{n} \) is a square. From here we see that if \( {t}_{n}\) is a square,

\( {t}_{4n(n+1)}=4\frac{n(n+1)}{2}.{(2n+1)}^{2} \) is also a square.

Hence if \({t}_{1}=1\) is a square then \( {t}_{8}=36\) is a square. And if \({t}_{8} \) is a square then \( {t}_{4*8(8+1)}={t}_{288}=144*289\) is a square.

Therefore,we get a sequence of infinite squares from here.

Saarthak Marathe - 2 years ago

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Brilliant solution!

Adarsh Kumar - 2 years ago

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Thank you. See my solution for 2nd question

Saarthak Marathe - 2 years ago

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Comment deleted Sep 25, 2015

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\((m,n) = (4,2)\)

Dev Sharma - 2 years ago

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\(2^m + 3^n = a^2\)

so \(m = 2x\) and \(n = 2y\)

then \(2^{2x} = a^2 - 3^{2y}\)

Dev Sharma - 2 years ago

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Comment deleted Sep 25, 2015

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@Adarsh Kumar By working on modulo 3 and 4 , we can get that....

Harsh Shrivastava - 2 years ago

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Comment deleted Sep 25, 2015

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@Adarsh Kumar I think that was a mere guess.

Nihar Mahajan - 2 years ago

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@Nihar Mahajan OR it was plug and check.

Mehul Arora - 2 years ago

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@Mehul Arora Or this margin is too small to contain his proof ......😛

Harsh Shrivastava - 2 years ago

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@Harsh Shrivastava

We have a large margin to fit in! :P

Mehul Arora - 2 years ago

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can anybody solve the rmo 2008 Maharashtra region problem 5

Devang Patil - 1 year, 12 months ago

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@Calvin Lin Sir, is it possible for you to close this note since we already have a part 2 for this note.

Brilliant Member - 2 years ago

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I don't see why this note should be locked. It is still valid for discussion, and adding of more comments / problems.

I might consider doing so if OP requests for it.

Calvin Lin Staff - 2 years ago

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First problem can be solved easily by using binomial therom

Sameer Pimparkhede - 2 years ago

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fermat's theorem, isn't it?

Raghav Rathi - 1 year, 10 months ago

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no need

Saarthak Marathe - 2 years ago

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All questions are done!

Saarthak Marathe - 2 years ago

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Could you please provide the solution for the \(6^{th}\) one?

Adarsh Kumar - 2 years ago

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Since no one has posted the solution for the 6th one yet, should I post it (solution given in the book)?

Brilliant Member - 1 year, 12 months ago

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@Brilliant Member Yes,please!

Adarsh Kumar - 1 year, 12 months ago

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For the \(2nd\) question, Case \(1\): If there is atleast one odd prime factor of \(m\),

Then by Euler's Totient function corolary, We can say that if prime \(p\) divides a number \(m\) then \(p-1\) divides \(\phi(m) \) . As \(p\) is odd, \(p-1\) is even, Therefore,\( 2|\phi(m) \).

Case \(2\): If there is no odd prime factor of \(m\),

Then \(m\) can be written as \(m={2}^{k} \) We can easily say that \( \phi(m)={2}^{k}-{2}^{k-1} \) Hence,\(2|\phi(m) \)

Saarthak Marathe - 2 years ago

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Comment deleted Sep 26, 2015

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Please refrain from posting questions here.You can give them to me and i will post them at the top mentioning that it has been given by you.That way,it is more accessible.

Adarsh Kumar - 2 years ago

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Comment deleted Sep 26, 2015

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@Brilliant Member Yes,please.I have posted it giving credit to you.BTW from where did you get this?

Adarsh Kumar - 2 years ago

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@Adarsh Kumar An excursion in Mathematics.

Brilliant Member - 2 years ago

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@Brilliant Member BTW who is the author? @Svatejas Shivakumar

Adarsh Kumar - 2 years ago

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@Brilliant Member Ok!Thanx!

Adarsh Kumar - 2 years ago

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6) \(3n + 1 = a^2\)

\(3n = a^2 - 1\)

\(3n = (a + 1)(a - 1)\)

so \(a + 1 = 3 so a = 2\) OR

\(a - 1 = 3 so a = 4\)

then \(n = 1 or n = 5\)

so there are two n...

Dev Sharma - 2 years ago

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Is this solution correct? I don't think so...

Harsh Shrivastava - 2 years ago

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@Harsh Shrivastava @Nihar Mahajan @Dev Sharma @Mehul Arora I am sorry but question 6 is wrong.It is satisfied fro more than one value of \(n\).It is my fault as i should have checked the problem before posting it.Sorry for the inconvenience caused.I am going to delete it.

Adarsh Kumar - 2 years ago

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@Adarsh Kumar P has to be a prime I told you that.

Mehul Arora - 2 years ago

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@Mehul Arora Ni you didn't,go and see the conversation.

Adarsh Kumar - 2 years ago

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@Adarsh Kumar Oh , sorry if I didn't. Please repost the problem and mention that p is prime.

Mehul Arora - 2 years ago

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@Mehul Arora Ohk.

Adarsh Kumar - 2 years ago

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@Adarsh Kumar I have a question. How can I give you? (not on slack)

Dev Sharma - 2 years ago

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@Dev Sharma Give it here.

Mehul Arora - 2 years ago

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@Mehul Arora If \(x^3 = x + 1\) then determine integer a,b,c \(x^7 = ax^2 + bx + c\)

Dev Sharma - 2 years ago

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@Dev Sharma Thanx for the question!Kindly delete this comment as i have posted it giving credit to you.

Adarsh Kumar - 2 years ago

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Comment deleted Sep 26, 2015

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yes. Thanks.

Dev Sharma - 2 years ago

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You assumed that 'n' is prime, which is not given in the question.....

Harsh Shrivastava - 2 years ago

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Yeah , correct.

Nihar Mahajan - 2 years ago

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@Adarsh Kumar In question 7, please mention that a,b and c are distinct. :)

Mehul Arora - 2 years ago

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done!

Adarsh Kumar - 2 years ago

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Comment deleted Sep 26, 2015

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\(n = 1\)

Dev Sharma - 2 years ago

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We would appreciate if you post a complete solution :)

Nihar Mahajan - 2 years ago

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@Mehul Arora Thanks for the problem :)

Nihar Mahajan - 2 years ago

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You're welcome @Nihar Mahajan

Mehul Arora - 2 years ago

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@Adarsh Kumar I believe that you can add a comment to the board, whenever you add a question. otherwise, it would be difficult to keep track.

You can add a comment like "Q3. is up" or "Next question!" so that people who are interested or have commented on this board get a notification, and they can check it out. Thanks.

Mehul Arora - 2 years ago

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Oh nice idea thanx!

Adarsh Kumar - 2 years ago

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Looks like giving a good idea gets me a downvote :3

Anyway, Glad to help :)

Mehul Arora - 2 years ago

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Comment deleted Sep 25, 2015

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@Adarsh Kumar Never blamed you for it ;)

Mehul Arora - 2 years ago

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Thanks!

Harsh Shrivastava - 2 years ago

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So, what are today's problems?

Swapnil Das - 2 years ago

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let us try 2nd question :

we can make following case and i am writing totient function as E,

Case 1 - when \(m\) is prime then E(m) = m - 1, which is even.

Case 2 - when \(m\) is even, clearly it would contain 2 so it even.

Dev Sharma - 2 years ago

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"Clearly it would contain 2" please elaborate.

Adarsh Kumar - 2 years ago

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I am sorry. I explained wrong.

Case 2. If m is even then m would be in form m = even . odd then we know that euler of odd number is even. So it would be even. Well, there would be one more, that is, even = even. Even

Dev Sharma - 2 years ago

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@Dev Sharma I believe i am not correctly understanding your method.Do you mean that \(m=2^{k}*(2a+1)\) where \(2^k\) is the even part of the number and \(2a+1\) the odd part?

Adarsh Kumar - 2 years ago

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@Adarsh Kumar yes but i am not confident about case 2. And is case 1 and 3 correct?

Dev Sharma - 2 years ago

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@Dev Sharma Could you explain after this in your case 2?Your case 1 is correct,i am not so sure about case 3,sorry.You could ask someone else.Actually there is a very simple way of proving this.If you want me to post i will.

Adarsh Kumar - 2 years ago

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@Adarsh Kumar Please post solution.

Dev Sharma - 2 years ago

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is it correct? If not, whats the mistake?

Dev Sharma - 2 years ago

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Case 3 - when \(m\) is odd (composite), then m would be divisible by any prime (3,5,7etc) and we know that totient function is multiplicative so, E(m) = E(any prime)E(left out)

also from case 1 we know E(prime) is even, so its even.

Dev Sharma - 2 years ago

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would it help?

The following results are by Fermet Little Theorm:

\(a^p = a modp\)

\(b^p = b modp\)

then \(a^p + b^p = a + b modp\)

now \((a + b)^p = a + b modp\)

so \((a + b)^p = a^p + b^p modp\)

Dev Sharma - 2 years ago

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Comment deleted Sep 24, 2015

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No , his solution is correct. Well there are two forms of Fermat's little theorem.

1) \(a^{p-1} \equiv 1 \pmod{p}\) where \(p\) is a prime and \(a\) is an integer such that \(gcd(a,p)=1\).

2) \(a^p \equiv a \pmod{p}\) where \(p\) is a prime and \(a\) is an integer.

You can see that if you use form (2) , there is no restriction saying \(gcd(a,p)=1\).

Nihar Mahajan - 2 years ago

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Yeah you are right it was pretty simple.Sorry!

Adarsh Kumar - 2 years ago

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Comment deleted Sep 26, 2015

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Comment deleted Sep 25, 2015

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Comment deleted Sep 26, 2015

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@Nihar Mahajan No not that,i thought it was not simple and i posted it.

Adarsh Kumar - 2 years ago

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Comment deleted Sep 26, 2015

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@Nihar Mahajan Good point!Ok i am removing it.

Adarsh Kumar - 2 years ago

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Comment deleted Sep 26, 2015

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Hey, I posted many RMO practice proof problems but there are 0 comment. (go to my profile to see them).

Dev Sharma - 2 years ago

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Thanks for \(Latex\)

Dev Sharma - 2 years ago

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For the \(8th\) question,

\(3n=(k-1)(k+1) \)

By the Prime factorisation principle, There is a unique prime factorisation for every number.

From here we can say that,

\( 3=k-1\) and \(k+1=n\)

Therefore,\( n=5\) is the only prime of this form.

Saarthak Marathe - 2 years ago

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