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. – Nihar Mahajan · 1 year, 1 month ago

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– Adarsh Kumar · 1 year, 1 month ago

Thanx for your solution.Log in to reply

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 · 1 year, 1 month ago

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– Ayush Verma · 9 months, 4 weeks ago

How u got this......explainLog in to reply

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 · 1 year, 1 month ago

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– Adarsh Kumar · 1 year, 1 month ago

No need of all that.See my solution.Log in to reply

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 · 1 year, 1 month ago

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– Nihar Mahajan · 1 year, 1 month ago

Is the first statement of your solution a lemma?Log in to reply

– Surya Prakash · 1 year, 1 month ago

Yes. It is a lemma. it is easy to proveLog in to reply

– Nihar Mahajan · 1 year, 1 month ago

Can you give me hint to prove it? Is it necessary to define a degree to the polynomials?Log in to reply

– Surya Prakash · 1 year, 1 month ago

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.Log in to reply

– Adarsh Kumar · 1 year, 1 month ago

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.Log in to reply

– Adarsh Kumar · 1 year, 1 month ago

Thanx for the solution.Log in to reply

– Anik Mandal · 1 year, 1 month ago

Thanks for the solution!Log in to reply

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.– Svatejas Shivakumar · 1 year agoLog in to reply

– Adarsh Kumar · 1 year ago

Really elegant one! Thanx for posting it!Log in to reply

– Svatejas Shivakumar · 1 year ago

Your welcome :)Log in to reply

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 · 1 year, 1 month ago

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– Adarsh Kumar · 1 year, 1 month ago

Brilliant solution!Log in to reply

– Saarthak Marathe · 1 year, 1 month ago

Thank you. See my solution for 2nd questionLog in to reply

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– Dev Sharma · 1 year, 1 month ago

\((m,n) = (4,2)\)Log in to reply

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

then \(2^{2x} = a^2 - 3^{2y}\) – Dev Sharma · 1 year, 1 month ago

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– Harsh Shrivastava · 1 year, 1 month ago

By working on modulo 3 and 4 , we can get that....Log in to reply

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– Nihar Mahajan · 1 year, 1 month ago

I think that was a mere guess.Log in to reply

– Mehul Arora · 1 year, 1 month ago

OR it was plug and check.Log in to reply

– Harsh Shrivastava · 1 year, 1 month ago

Or this margin is too small to contain his proof ......😛Log in to reply

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can anybody solve the rmo 2008 Maharashtra region problem 5 – Devang Patil · 1 year 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. – Svatejas Shivakumar · 1 year ago

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I might consider doing so if OP requests for it. – Calvin Lin Staff · 1 year ago

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First problem can be solved easily by using binomial therom – Sameer Pimparkhede · 1 year, 1 month ago

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– Raghav Rathi · 10 months, 3 weeks ago

fermat's theorem, isn't it?Log in to reply

– Saarthak Marathe · 1 year, 1 month ago

no needLog in to reply

All questions are done! – Saarthak Marathe · 1 year, 1 month ago

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– Adarsh Kumar · 1 year ago

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

– Svatejas Shivakumar · 1 year ago

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

– Adarsh Kumar · 1 year ago

Yes,please!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. – Saarthak Marathe · 1 year, 1 month 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 · 1 year, 1 month ago

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– Adarsh Kumar · 1 year, 1 month ago

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.Log in to reply

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– Adarsh Kumar · 1 year, 1 month ago

Yes,please.I have posted it giving credit to you.BTW from where did you get this?Log in to reply

– Svatejas Shivakumar · 1 year, 1 month ago

An excursion in Mathematics.Log in to reply

@Svatejas Shivakumar – Adarsh Kumar · 1 year, 1 month ago

BTW who is the author?Log in to reply

– Adarsh Kumar · 1 year, 1 month ago

Ok!Thanx!Log in to reply

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 · 1 year, 1 month ago

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– Harsh Shrivastava · 1 year, 1 month ago

Is this solution correct? I don't think so...Log in to reply

@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 · 1 year, 1 month ago

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@Adarsh Kumar P has to be a prime I told you that. – Mehul Arora · 1 year, 1 month ago

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– Adarsh Kumar · 1 year, 1 month ago

Ni you didn't,go and see the conversation.Log in to reply

– Mehul Arora · 1 year, 1 month ago

Oh , sorry if I didn't. Please repost the problem and mention that p is prime.Log in to reply

– Adarsh Kumar · 1 year, 1 month ago

Ohk.Log in to reply

– Dev Sharma · 1 year, 1 month ago

I have a question. How can I give you? (not on slack)Log in to reply

– Mehul Arora · 1 year, 1 month ago

Give it here.Log in to reply

– Dev Sharma · 1 year, 1 month ago

If \(x^3 = x + 1\) then determine integer a,b,c \(x^7 = ax^2 + bx + c\)Log in to reply

– Adarsh Kumar · 1 year, 1 month ago

Thanx for the question!Kindly delete this comment as i have posted it giving credit to you.Log in to reply

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– Dev Sharma · 1 year, 1 month ago

yes. Thanks.Log in to reply

– Harsh Shrivastava · 1 year, 1 month ago

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

– Nihar Mahajan · 1 year, 1 month ago

Yeah , correct.Log in to reply

@Adarsh Kumar In question 7, please mention that a,b and c are distinct. :) – Mehul Arora · 1 year, 1 month ago

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– Adarsh Kumar · 1 year, 1 month ago

done!Log in to reply

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– Dev Sharma · 1 year, 1 month ago

\(n = 1\)Log in to reply

– Nihar Mahajan · 1 year, 1 month ago

We would appreciate if you post a complete solution :)Log in to reply

@Mehul Arora Thanks for the problem :) – Nihar Mahajan · 1 year, 1 month ago

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@Nihar Mahajan – Mehul Arora · 1 year, 1 month ago

You're welcomeLog in to reply

@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 · 1 year, 1 month ago

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– Adarsh Kumar · 1 year, 1 month ago

Oh nice idea thanx!Log in to reply

Anyway, Glad to help :) – Mehul Arora · 1 year, 1 month ago

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– Mehul Arora · 1 year, 1 month ago

Never blamed you for it ;)Log in to reply

Thanks! – Harsh Shrivastava · 1 year, 1 month ago

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So, what are today's problems? – Swapnil Das · 1 year, 1 month 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 · 1 year, 1 month agoLog in to reply

– Adarsh Kumar · 1 year, 1 month ago

"Clearly it would contain 2" please elaborate.Log in to reply

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 · 1 year, 1 month ago

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– Adarsh Kumar · 1 year, 1 month ago

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?Log in to reply

– Dev Sharma · 1 year, 1 month ago

yes but i am not confident about case 2. And is case 1 and 3 correct?Log in to reply

– Adarsh Kumar · 1 year, 1 month ago

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.Log in to reply

– Dev Sharma · 1 year, 1 month ago

Please post solution.Log in to reply

– Dev Sharma · 1 year, 1 month ago

is it correct? If not, whats the mistake?Log in to reply

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 · 1 year, 1 month 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 · 1 year, 1 month ago

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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 · 1 year, 1 month ago

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– Adarsh Kumar · 1 year, 1 month ago

Yeah you are right it was pretty simple.Sorry!Log in to reply

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– Adarsh Kumar · 1 year, 1 month ago

No not that,i thought it was not simple and i posted it.Log in to reply

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– Adarsh Kumar · 1 year, 1 month ago

Good point!Ok i am removing it.Log in to reply

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– Dev Sharma · 1 year, 1 month ago

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

\(Latex\)– Dev Sharma · 1 year, 1 month agoLog in to reply