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).

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TopNewest5) 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.

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Thanx for your solution.

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

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How u got this......explain

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

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No need of all that.See my solution.

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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.

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Is the first statement of your solution a lemma?

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

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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.

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Thanx for the solution.

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Thanks for the solution!

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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.Log in to reply

Really elegant one! Thanx for posting it!

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Your welcome :)

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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.

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Brilliant solution!

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Thank you. See my solution for 2nd question

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Comment deleted Sep 25, 2015

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

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

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

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

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can anybody solve the rmo 2008 Maharashtra region problem 5

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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.

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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.

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First problem can be solved easily by using binomial therom

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fermat's theorem, isn't it?

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no need

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All questions are done!

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

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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)?

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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,\(2|\phi(m) \)

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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.

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@Svatejas Shivakumar

BTW who is the author?Log in to reply

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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...

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

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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.

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

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Comment deleted Sep 26, 2015

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yes. Thanks.

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

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Yeah , correct.

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

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done!

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Comment deleted Sep 26, 2015

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\(n = 1\)

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We would appreciate if you post a complete solution :)

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@Mehul Arora Thanks for the problem :)

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You're welcome @Nihar Mahajan

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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.

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Oh nice idea thanx!

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

Anyway, Glad to help :)

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Comment deleted Sep 25, 2015

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Thanks!

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So, what are today's problems?

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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.Log in to reply

"Clearly it would contain 2" please elaborate.

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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

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is it correct? If not, whats the mistake?

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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.

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

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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 \(gcd(a,p)=1\).

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Yeah you are right it was pretty simple.Sorry!

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

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Thanks for

\(Latex\)Log in to reply

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.

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