Open Proof Contest 2

Hello Everysinx6n\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 2nd2^{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.Prove that 102n+11 is never a perfect square. For nN1. \text{Prove that } 10^{2n + 1} - 1 \text{ is never a perfect square. For } n \in N

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

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

4.Let d be any positive integer not equal to 2,5,or13. Show that one can find distinct a,b in the set2,5,13,d such that  ab1is not a perfect square.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.Prove that if n4thenn,n+2,n+4 can’t all be primes.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

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

Please comment and inform me if willing to participate.

Theme : Perfect Square

Note by Rajdeep Dhingra
4 years, 7 months ago

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1 vote

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Comments

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Nice idea of starting a website !

A Former Brilliant Member - 4 years, 7 months ago

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Thanks ! Will you participate or apply.

Rajdeep Dhingra - 4 years, 7 months ago

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

A Former Brilliant Member - 4 years, 7 months ago

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Can you clarify what question 3 means?

Agnishom Chattopadhyay Staff - 4 years, 7 months ago

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He means that , if LCM(a,b)=GCD(a,b) , then prove a=±ba=\pm b

A Former Brilliant Member - 4 years, 7 months ago

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

Rajdeep Dhingra - 4 years, 7 months ago

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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 - 4 years, 7 months ago

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yes you can. Please reshare the note . Also ask more people to participate.

Thanking you

Rajdeep

Rajdeep Dhingra - 4 years, 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 - 4 years, 7 months ago

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Hello,

When you say natural numbers, do you mean positive integers?

Daniel Liu - 4 years, 7 months ago

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Yes

Rajdeep Dhingra - 4 years, 7 months ago

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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 - 4 years, 7 months ago

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Yes you can

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

Thanks for participating

Rajdeep Dhingra - 4 years, 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 - 4 years, 7 months ago

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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.

Rajdeep Dhingra - 4 years, 7 months ago

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@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 - 4 years, 7 months ago

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Lemma 1

Solution to 3-5

Solution to 1

solution to 2

Results

Rajdeep Dhingra - 4 years, 6 months ago

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The solutions have been nicely written out :)

Looking forward to the final document .

A Former Brilliant Member - 4 years, 6 months ago

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Results are out !

Rajdeep Dhingra - 4 years, 6 months ago

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@Rajdeep Dhingra Ok , I just saw it .

A Former Brilliant Member - 4 years, 6 months ago

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I see that you posted my Lemma 1 and Solutions 1,2. I'm honoured for that. :)

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 - 4 years, 6 months ago

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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

Rajdeep Dhingra - 4 years, 6 months ago

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@Rajdeep Dhingra 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. :)

Prasun Biswas - 4 years, 6 months ago

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@Prasun Biswas Well the deadline will be 15th April. So can you ?

Rajdeep Dhingra - 4 years, 6 months ago

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Results are Out

Rajdeep Dhingra - 4 years, 6 months ago

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

Mehul Arora - 4 years, 6 months ago

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100100 is a multiple of 44 and so any arbitrary positive integral power of it will also be a multiple of 44. Hence, we have,

100n0(mod4) , nZ+100^n\equiv 0\pmod{4}~,~n\in\mathbb{Z^+}

Now, if you multiply any integer with a multiple of 44, the resulting product will also be a multiple of 44. Hence, we have,

100n×100(mod4)100^n\times 10\equiv 0\pmod{4}

Now, using the properties of modular arithmetic, when you subtract (1)(-1) from a multiple of 44, you get a remainder of (1)(-1), or by cyclicity of remainders, we get a remainder of (1+4)=3(-1+4)=3 modulo 44. Mathematically,

100n×10101(1)(1+4)3(mod4)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 - 4 years, 6 months ago

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Damn! I was a bit careless. Thanks for clearing my doubt !! @Prasun Biswas . You rock!

Mehul Arora - 4 years, 6 months ago

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I think Prasun has already cleared your doubt.

Rajdeep Dhingra - 4 years, 6 months ago

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