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Open Proof Contest 2

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

Note by Rajdeep Dhingra
2 years, 11 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 $${100}^{n}$$ *10 -1 is congruent to 0-1 is congruent to 3 (mod 4)?

- 2 years, 10 months ago

$$100$$ is a multiple of $$4$$ and so any arbitrary positive integral power of it will also be a multiple of $$4$$. Hence, we have,

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

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. :)

- 2 years, 10 months ago

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

- 2 years, 10 months ago

- 2 years, 10 months ago

- 2 years, 10 months ago

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. $$\ddot\frown$$

- 2 years, 10 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. $$\ddot\smile$$

- 2 years, 10 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. :)

- 2 years, 10 months ago

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

- 2 years, 10 months ago

The solutions have been nicely written out :)

Looking forward to the final document .

- 2 years, 10 months ago

Results are out !

- 2 years, 10 months ago

Ok , I just saw it .

- 2 years, 10 months ago

@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

- 2 years, 11 months ago

@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)

- 2 years, 11 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.

- 2 years, 11 months ago

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)?

- 2 years, 11 months ago

Yes you can

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

Thanks for participating

- 2 years, 11 months ago

Hello,

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

- 2 years, 11 months ago

Yes

- 2 years, 11 months ago

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!

- 2 years, 11 months ago

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

Thanking you

Rajdeep

- 2 years, 11 months ago

@Francis Gerard Magtibay
Ignore if not sent.
P.S - Are you sending ?

- 2 years, 11 months ago

Can you clarify what question 3 means?

- 2 years, 11 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.

- 2 years, 11 months ago

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

- 2 years, 11 months ago

Nice idea of starting a website !

- 2 years, 11 months ago

Thanks ! Will you participate or apply.

- 2 years, 11 months ago

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

- 2 years, 11 months ago

Comment deleted Mar 15, 2015

Since I have my board exams going on, I don't think I'll be able to participate for the time being (I'll try to make up some free time though, if possible). After 23rd March, I'll be active like before.

- 2 years, 11 months ago

I think that I'll be active like never before !!

- 2 years, 11 months ago

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

- 2 years, 11 months ago