Hello Every\(\frac{\sin{x}}{6n}\).

This is the second part of the contest started by me. The first one was cancelled as many people insisted on it as there exams were going .They said to release its second part.Here is the \(2^{nd}\) part. The rules are the same as the first one. You need to mail the PDF to opencontestproofs@gmail.com. To make a pdf you can go to Latex PDF. Also you can send a snip of the solution after typing it in DAUM.

Here are the questions

\(1. \text{Prove that } 10^{2n + 1} - 1 \text{ is never a perfect square. For } n \in N\)

\( 2. \text{Prove that no number in this sequence } 11,111,1111,11111,........... \\ \text{ is a perfect square} \)

\(3. If \quad [a,b] = (a,b) . \text{ Then show that } a = \pm b.\) [a,b] means LCM. (a,b) means GCD.

\(4. \text{Let } d \text{ be any positive integer not equal to } 2,5, or 13. \\ \text{ Show that one can find distinct } a,b \text{ in the set} 2,5,13,d \text{ such that }\ ab-1 \text{is not a perfect square.} \)

\(5. \text{Prove that if } n \ge 4 \quad then \quad n , n+2,n+4 \text{ can't all be primes.}\)

Please Reshare so more can participate.

Open for all.

Ended

\[\textbf{Solutions link in the comment}\]

Please comment and inform me if willing to participate.

Theme : Perfect Square

## Comments

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TopNewest@Rajdeep Dhingra , I have a doubt In the solution of the first Question. In the Second step, How did you conclude that \({100}^{n}\) *10 -1 is congruent to 0-1 is congruent to 3 (mod 4)? – Mehul Arora · 1 year, 7 months ago

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\[100^n\equiv 0\pmod{4}~,~n\in\mathbb{Z^+}\]

Now, if you multiply any integer with a multiple of \(4\), the resulting product will also be a multiple of \(4\). Hence, we have,

\[100^n\times 10\equiv 0\pmod{4}\]

Now, using the properties of modular arithmetic, when you subtract \((-1)\) from a multiple of \(4\), you get a remainder of \((-1)\), or by cyclicity of remainders, we get a remainder of \((-1+4)=3\) modulo \(4\). Mathematically,

\[100^n\times 10 - 1\equiv 0-1\equiv (-1)\equiv (-1+4)\equiv 3\pmod{4}\]

If you have any more doubts, reply to this comment. I'll reply to it as soon as possible. :) – Prasun Biswas · 1 year, 7 months ago

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@Prasun Biswas . You rock! – Mehul Arora · 1 year, 7 months ago

Damn! I was a bit careless. Thanks for clearing my doubt !!Log in to reply

– Rajdeep Dhingra · 1 year, 7 months ago

I think Prasun has already cleared your doubt.Log in to reply

Results are Out – Rajdeep Dhingra · 1 year, 7 months ago

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Lemma 1

Solution to 3-5

Solution to 1

solution to 2

Results – Rajdeep Dhingra · 1 year, 7 months ago

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P.s - I figured out solutions to 3 and 5 a few days ago and was planning to submit them today since you earlier said that the deadline was until 28th March which is today. I'm sad to see that you changed the deadline without notice. \(\ddot\frown\) – Prasun Biswas · 1 year, 7 months ago

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– Rajdeep Dhingra · 1 year, 7 months ago

Sorry for ending it a day before. Actually I saw that no one has submitted solutions in the last days so I had ended the contest. In OPC 3 just inform me that you are going to submit solutions. \(\ddot\smile\)Log in to reply

– Prasun Biswas · 1 year, 7 months ago

Ah, okay. By the way, I won't be active for some time since my PC is damaged again. I'm writing this from a cyber cafe. I'm unsure whether I'll take part in OPC 3 but if I do, I'll let you know. :)Log in to reply

– Rajdeep Dhingra · 1 year, 7 months ago

Well the deadline will be 15th April. So can you ?Log in to reply

Looking forward to the final document . – Azhaghu Roopesh M · 1 year, 7 months ago

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– Rajdeep Dhingra · 1 year, 7 months ago

Results are out !Log in to reply

– Azhaghu Roopesh M · 1 year, 7 months ago

Ok , I just saw it .Log in to reply

@Jason Martin

Sorry sir

But your solutions are unclear. Could you send me again and check your mail for the email I sent to you.

Thanking You

Rajdeep – Rajdeep Dhingra · 1 year, 7 months ago

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@Rajdeep Dhingra ... Thanks to all your support, I got 2 of these 5 problems... I hopefully will be writing their proofs by the 20th because I have some workshops till then.... Thanks to all your support.

Just curious:- Did you make and/or solve all these problems by yourself? (I am pretty sure you were able to coz you are one of the best by age I know) – Mehul Arora · 1 year, 7 months ago

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– Rajdeep Dhingra · 1 year, 7 months ago

Of course bro, Solve them. Well I made ques 4 and 1 and 3 , Also I solved them all and will be launching a solution manual on 29th March.Log in to reply

By the way, can I submit my solution in parts? That is to say, can I submit my solution to two or three of your problems now and the rest of them later?

Also, where do I submit the solution file(s)? – Prasun Biswas · 1 year, 7 months ago

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The solution files have to be emailed to the email mentioned in this note.

Thanks for participating – Rajdeep Dhingra · 1 year, 7 months ago

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Hello,

When you say natural numbers, do you mean positive integers? – Daniel Liu · 1 year, 7 months ago

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– Rajdeep Dhingra · 1 year, 7 months ago

YesLog in to reply

Hi, I'd like to submit my proofs but I'm not familiar with LaTeX. Will you accept submitted entries in MS Word? I'll send both the original Word file and a compilation of screenshots so you have a read-only copy, if you want it that way. Thanks! – Francis Gerard Magtibay · 1 year, 7 months ago

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Thanking you

Rajdeep – Rajdeep Dhingra · 1 year, 7 months ago

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@Francis Gerard Magtibay

I haven't received your solutions. Just informing.

Ignore if not sent.

P.S - Are you sending ? – Rajdeep Dhingra · 1 year, 7 months ago

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Can you clarify what question 3 means? – Agnishom Chattopadhyay · 1 year, 7 months ago

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– Rajdeep Dhingra · 1 year, 7 months ago

It means when the lcm of 2 numbers is equal to hcf of those 2 numbers . Then find the relation b\w those numbers.Log in to reply

– Azhaghu Roopesh M · 1 year, 7 months ago

He means that , if LCM(a,b)=GCD(a,b) , then prove \(a=\pm b\)Log in to reply

Nice idea of starting a website ! – Azhaghu Roopesh M · 1 year, 7 months ago

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– Rajdeep Dhingra · 1 year, 7 months ago

Thanks ! Will you participate or apply.Log in to reply

– Azhaghu Roopesh M · 1 year, 7 months ago

Neither for the time being , sorry about that ! I'll surely be free after 24 May , I'll see after that :)Log in to reply

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And I have already liked and reshared your note. :) – Prasun Biswas · 1 year, 7 months ago

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– Azhaghu Roopesh M · 1 year, 7 months ago

I think that I'll be active like never before !!Log in to reply

– Rajdeep Dhingra · 1 year, 7 months ago

Thanks, Due to exams I have kept the deadline on 28th March.Log in to reply