BR Maths Corner-1

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Amber G.
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Re: BR Maths Corner-1

Postby Amber G. » 02 Mar 2011 03:00

You may already have it, but
http://www-history.mcs.st-andrews.ac.uk/HistTopics/Bakhshali_manuscript.html
may be of help wrt to Bakhshali manuscript. Original, I think, is in the Library at the University of Oxford but there are many online pictures.
eg:
Image


As to etymology of zero: per any standard reference eg
[url](http://www.merriam-webster.com/dictionary/zero[/url]) one sees:
( "zero" came via French zéro from Venetian zero, which (together with cipher) came via Italian zefiro from Arabic صفر, (ṣifr) which came from शुन्य )
(Arabs changed the Sanskrit word ‘Shunya'' to ‘sifr' around 10th century In 12th century, Italians (mathematician Fibonacci studied Arabian algebra) introduced the Hindu-Arabic numerals in Italy (BTW Arabs called the numbers 'Hindsa' = from hindus). In Italy, 'sifr' was Latinized to ‘Zephirum'...which over time became zero. (BTW in Germany Nemaririus retained the original but modified it to 'cifra' - in England it became 'cipher' ..in early period the zero was looked upon as a secret/mystifying sign by the common people..hence words like ‘decipher' reveals the enigma associated with it)

One of the oldest manuscript which used the word 'shunya' is Lokavibhaaga (Jain)..(which btw was used as a text-book by Al-Khwarizmi centuries later! ...the rough translation from an Arabic book of that time goes something like "We must know that the Indians have a most subtle talent and all other races yield to them in arithmetic and geometry and the other liberal arts. And this is clear in the figures with which they are able to designate each and every degree of each order (of numbers)"
Hth

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Re: BR Maths Corner-1

Postby Multatuli » 22 Mar 2011 23:26

Would it be okay if I posted some problems in here fom "A course of Pure Mathematics" and other books? Problems I could not solve by myself, in all these years?

There are some really beautiful formulae I never could prove. Would be nice to see how it's done.

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Re: BR Maths Corner-1

Postby Amber G. » 27 Mar 2011 05:25

I am sure that it is okay to post.

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Re: BR Maths Corner-1

Postby SaiK » 27 Mar 2011 05:30

Amber ji, any good text books on discrete math that you recommend.

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Re: BR Maths Corner-1

Postby ramana » 10 May 2011 01:55

Hindu article

Welcome to Wiki Maths

[quote]
Welcome to WikiMaths, home of hard sums

Matt Parker

The Polymath Project allows unprecedented numbers of people to work on the same problem, hopefully solving conundrums more quickly.

Mathematicians are not known as a social bunch, but a new “WikiMaths” project is allowing anyone to join in their cutting-edge research. A study into the effectiveness of the world's first virtual mathematics project will be released this week.

It all started in 2009, when Cambridge mathematician Tim Gowers wrote about the possibility of an open online group allowing unprecedented numbers of people to work on the same problem, hopefully solving conundrums much more quickly. He suggested the “Hales-Jewett theorem” as a good first target.

Analogous to a complicated game of noughts and crosses played on a 4x4 cube in five dimensions, the theorem shows how many squares you would need to block to make it impossible to complete any straight lines. On a 3x3 grid, you can do this by blocking three squares; in five dimensions, things are a bit more complicated.

Truly collaborative

This theorem had already been proven, but the solution was long and complicated and no one had found a much-needed basic proof.

Contributions poured in — a staggering 1,228 significant comments across 14 blog posts with 39 people providing meaningful contributions. Within six weeks the answer had been found. It was published under the collective pseudonym “DHJ Polymath”.

But was the process truly collaborative? Researchers at Carnegie Mellon University in Pittsburgh, think so. Much of the work was done by professional mathematicians, but a number of smaller, vital contributions came from those without serious credentials.

The 39 contributors to the Hales-Jewett theorem solution ranged from the world's top mathematicians to secondary school maths teachers. Several seminal ideas came from inexperienced mathematicians. Which all means that the exercise could redefine who is considered a mathematician — and offer new insight into unsolved problems.

The researchers are presenting their results in Vancouver next week, while the “Polymath Project”, as it now known, continues to work on seven different problems with more than 5,000 comments from 275 unique contributors. Why not join in at polymathprojects.org? — © Guardian Newspapers Limited, 2011

[/url]

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Re: BR Maths Corner-1

Postby Amber G. » 21 May 2011 20:09

SaiK wrote:Amber ji, any good text books on discrete math that you recommend.

SaiK - I do not have any particular book in mind , but if you do not know already, one of the best resources is IIT online courses.. (check out google..)
For example, this (or related ones) from IIT ..
http://freevideolectures.com/Course/2661/Discrete-Mathematical-Structures/3

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Re: BR Maths Corner-1

Postby Amber G. » 21 May 2011 20:22

Okay, here are two nice simple :) problems from a current national math Olympiads.. (Name of the country is omitted to make googling non-trivial :).. and the exam has passed so putting solution here is okay)

P1: (Problem is for high school kids)

Given, x, y, z are real positive numbers.. and
x+y+z = 3/670
and
1/x + 1/y + 1/z = 2010

Find, x, y, z
(All possible values, of course .. not just one or two...)

*****
P2

Problem from Junior Math Olympiad ...(For kids not yet in high-school)
Find (with proof) all n such that 2^n+12^n+2011^n is a perfect square..
(n is a natural number)

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Re: BR Maths Corner-1

Postby Amber G. » 21 May 2011 21:05

BTW, the very first problem from me in this br math corner:
http://forums.bharat-rakshak.com/viewtopic.php?f=2&t=4201&p=514596&hilit=Prove+that+100+math+corner#p514596
was (slightly different language, but exactly the same concept) actually appeared in one of the prestigious "entrance" exam in this year (recently) India.. 8)

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Re: BR Maths Corner-1

Postby rsingh » 22 May 2011 01:07

Bah could nat solve any of the problems. Do you have some real problems which reflect day to day dealings at Madaesahs :roll: ................I mean abdul chacha had 5 wife 3 bakri and 4 klashnikovs.........how much he property he has.

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Re: BR Maths Corner-1

Postby Amber G. » 30 May 2011 08:25

Recently announced, here is India IMO team for 2011 - Wish them best.
- Akashdeep (West Bengal)
- Akashnil(12th; West Bengal)
- Debdyuti Banerjee(or Chandan)(11th, West Bengal)
- Indraneel Kasmalkar(12th, Pune)
- Prafulla Sushil dhariwal(10th, Pune)
- Mrudul Thatte(10th, Pune)

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Re: BR Maths Corner-1

Postby Shrinivasan » 30 May 2011 10:44

All the best kids, these boys and girls are the future of our nation!!!

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Re: BR Maths Corner-1

Postby S.G.Dev » 30 May 2011 11:03

Amber G. wrote:Okay, here are two nice simple :) problems from a current national math Olympiads.. (Name of the country is omitted to make googling non-trivial :).. and the exam has passed so putting solution here is okay)

P1: (Problem is for high school kids)

Given, x, y, z are real positive numbers.. and
x+y+z = 3/670
and
1/x + 1/y + 1/z = 2010

Find, x, y, z
(All possible values, of course .. not just one or two...)


Besides the obvious x=y=z=1/670, what are the other solutions and what's the process for getting them? There are 2 equations with 3 unknowns.

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Re: BR Maths Corner-1

Postby chilarai » 30 May 2011 14:49

S.G.Dev wrote:
Amber G. wrote:Okay, here are two nice simple :) problems from a current national math Olympiads.. (Name of the country is omitted to make googling non-trivial :).. and the exam has passed so putting solution here is okay)

P1: (Problem is for high school kids)

Given, x, y, z are real positive numbers.. and
x+y+z = 3/670
and
1/x + 1/y + 1/z = 2010

Find, x, y, z
(All possible values, of course .. not just one or two...)


Besides the obvious x=y=z=1/670, what are the other solutions and what's the process for getting them? There are 2 equations with 3 unknowns.



Hint :: look at Cauchy-Schwarz inequality

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Re: BR Maths Corner-1

Postby Amber G. » 30 May 2011 22:28

chilarai wrote:
Hint :: look at Cauchy-Schwarz inequality

Yep, Very Nice!!

But there can also be simpler, in my opinion, ways to do it.

One way is to use rsingh's example, and see why one can solve it intuitively:
Let the speed of three bakris of Abdul chacha be x,y,z: And they ran in a relay race..

Case 1: Each bakri ran for 1 hour each, total distance traveled = 3/670 (mile)
Case 2: Eack bakri ran for 1 mile each ... Total time taken = 2010 hours

What is the speed of each Bakri?

(Think about it, Hint: Not very hard, once you get the right idea)
(Okay, these bakris are real slow -- make them kachuaa- answer will not change)

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Re: BR Maths Corner-1

Postby SaiK » 31 May 2011 00:52

Thanks for the links Amber. salute!

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Re: BR Maths Corner-1

Postby Amber G. » 31 May 2011 02:28

Got an e-mail... why our DDM has to be so silly :roll:

Here is a news story about Indian IMO team: (March story)
Math crown for Bokaro wizards
Bokaro, March 24: What’s in a number? The sum of all successes, says DPS.

Five students from this elite city school have not just cracked the International Mathematics Olympiad, but have also bagged the winner and runners-up titles at the meet, which saw some 8,000 participants from across the globe....

Never mind that IMO is in July in Amsterdam. :-o
(The same story has appeared in few other newspapers too)

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Re: BR Maths Corner-1

Postby Najunamar » 31 May 2011 04:09

The second problem posted by Amberji-

Using modulo 2 on both sides, it is evident that it is an odd square, since 2 and 12 are both even and 2011 isn't. Also since every square is 0mod5 or +/- 1 mod 5, and 2011 is 1 mod 5, we get the result that n is 1 mod 4, not just odd but skips evey other odd number. Now using modulo 4, every square is either 0mod4 or 1mod4, we know RHS is 0 or 1 mod 4, but 2011 is -1 mod 4, with 2^n being either 2 mod 4 for n=1 or 0 mod 4 for all other higher powers of n that are 1 mod 4, 12 is 0mod 4, hence for all n>1 LHS is -1^n mod 4 which cannot be +1 mod 4 since n is odd, Hence no solution other than n = 1 which equates to 45^2 is possible.

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Re: BR Maths Corner-1

Postby Amber G. » 31 May 2011 07:52

Also since every square is 0mod5 or +/- 1 mod 5 {Agree} , and 2011 is 1 mod 5, we get the result that n is 1 mod 4, not just odd but skips every other odd number.

Why, I am not very clear, why can't n be 2 mod 4? Because LHS at mod 5; 2^2+12^2+2011^2 == 2^2+2^2+1
which is 9 (or -1) mod 5 which could be a perfect square... :-?

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Re: BR Maths Corner-1

Postby Najunamar » 31 May 2011 23:57

Must have left out this portion, taking mod 3 on both sides shows in order to be not 2 mod 3 (2 being a quadratic non residue of 3) n has to be odd.

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Re: BR Maths Corner-1

Postby Amber G. » 01 Jun 2011 02:46

Yes, thanks. Nice solution.

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Re: BR Maths Corner-1

Postby Amber G. » 01 Jun 2011 03:06

S.G.Dev wrote:Besides the obvious x=y=z=1/670, what are the other solutions and what's the process for getting them? There are 2 equations with 3 unknowns.

Hi, as one might have guessed, there is additional information.. that is all (x,y,z) are positive.

As Chilarai suggested Cauchy-Schwarz inequality is one way. I hope he does not mind, for those who may not be familiar with this, it is simply (in 3 dim, special case) (see wiki)

(p^2+q^+r^2)(a^2+b^2+c^2) >= (pa+qb+rc)^2
Equality holds only when p/a = q/b =r/c ..
Here, if one takes p^2 = x, q^2=y, z^2=z and a^2=1/x, b^2=1/y,c^2=1/c one sees the equality holds... hence x=y=z.. rest is simple.


Another way to look at is to notice that both arithmetic mean (of x,y,z) and harmonic mean are equal (1/670), this can happen only if all are equal.

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Re: BR Maths Corner-1

Postby chilarai » 02 Jun 2011 06:56

Amber G. wrote:...

As Chilarai suggested Cauchy-Schwarz inequality is one way. I hope he does not mind, for those who may not be familiar with this, it is simply (in 3 dim, special case) (see wiki)
..


oh no issues at all . its not like i knew cauchy schwarz inequality before this problem.
in this particular problem , i only managed to reach the state of

(x+y+z)(1/x+1/y+1/z) = 9

and then i was lost , so i googled for (x+y+z)(1/x+1/y+1/z) , (a + b + c)(1/a + 1/b + 1/c) and
(p + q + r) (1/p + 1/q + 1/r ) ..and somehow one of these landed me to cauchy schwarz inequality.
:twisted:

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Re: BR Maths Corner-1

Postby Amber G. » 02 Jun 2011 23:45

Interesting!!!
After reading above, I tried google too..Here
Get the right answer! Jai Ho google ki!
Links I get are:<Top of the list> :)!
and <this>
(exe 3 - some nice discussion about the problem ) (Exam where it appeared)

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Re: BR Maths Corner-1

Postby chilarai » 17 Jun 2011 08:03

Hope not posted earlier

for those who likes programming and maths , there are some 300+ nice programming problems to solve at

http://projecteuler.net/

you may choose any programming language ( some masochists have used even assembly** to solve these , but personally i find haskell is just nice for these kind of problems )

most programs should run in less than one minute in any decent CPU if not your implementation is prolly inefficient !


**once you solve a problem you can access the forum where others may have posted their solutions

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Re: BR Maths Corner-1

Postby vina » 17 Jun 2011 11:19

Amber G. wrote:Okay, here are two nice simple :) problems from a current national math Olympiads.. (Name of the country is omitted to make googling non-trivial :).. and the exam has passed so putting solution here is okay)

P1: (Problem is for high school kids)

Given, x, y, z are real positive numbers.. and
x+y+z = 3/670
and
1/x + 1/y + 1/z = 2010

Find, x, y, z
(All possible values, of course .. not just one or two...)


Multiplying the two we get
(x+y+z)(1/x + 1/y + 1/z) = 9
after simplifying and rearranging we got
(x/y + y/x) + (z/x + z/x) + (y/z + z/y) = 6
Now since we know that (a + 1/a) >=2 and since all of x, y, z > 0 , the only way this can be satisfied is when each of the terms within the brackets = 2 and from that it is easy to get that x=y=z is the only answer.

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Re: BR Maths Corner-1

Postby vina » 17 Jun 2011 12:27

Amber G. wrote:P2

Problem from Junior Math Olympiad ...(For kids not yet in high-school)
Find (with proof) all n such that 2^n+12^n+2011^n is a perfect square..
(n is a natural number)


My answer is not an elegant "Proof", but rather a Paki like "Broof" , still ok I guess, since you said it is junior high level .

n =1 is the only possible answer.
Plugging in 0 for n, doesn't work. n = 1 works.

ie 2^1 + 12^1+2011^1 = 2025 = 45^2

Now
2^n + 12^n + 2011^n <= (2+12+2011)^n = 2025^n = (2011 + 14)^n

If we try expanding (2011 + 14)^n = (2011)^n + 14^n + positive terms

Re arranging a bit we have
2^n + 12^n + 2011^n - 2011^n - 14^n >= 0

ie 2^n + 12^n - 14^n >=0

Now n = 1 is the only solution . For all n > 1 it is easy to see that LHS becomes negative (the function is decreasing with increasing n ) ie 14^n = (12+2)^n >= 12^n + 2^n + positive terms > 12^n + 2^n


Okay.. There is an error in the above workings. I just punched it off since I got busy. All the same, let me try to "Imbroove" the "Broof" and try to make it a dignified "Proof"

We have 2^n + 12^n +2011^n to be a square must be equal to (2011+ 2k)^n .. The LHS is odd, hence it should be an odd number that is raised to power n.

2^n + 12^n +2011^n = (2011 + 2k) ^n

Expanding the RHS as 2011^n + (2k)^n + (positive terms of lower order of n)

Since the LHS only powers of n, collectively the positive terms on the RHS of lower order must be zero for equality to hold

So rearranging, we have
2^n + 12^n +2011^n - 2011^n - (2k)^n = 0

or 2^n + 12^n - (2k)^n = 0

For natural numbers of n, putting n = 1, the min value of k = 7.

For values > 1, 2^n + 12^n - (2k)^n is less than 0 for all values of n (as discussed in the previous para). So k = 7 .

With k = 7, the only value that will satisfy the equation is n = 1 and that is the only solution for the problem!

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Re: BR Maths Corner-1

Postby SwamyG » 17 Jun 2011 21:10

http://www.artofproblemsolving.com/Store/index.php?

Anybody have any experiences in using the books from the "Art of Problem Solving" book store? My son enters the MathCounts competition, and I was wondering on the utility of these books.

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Re: BR Maths Corner-1

Postby Amber G. » 17 Jun 2011 22:53

Sirji - confused only...:) (See notes inside)
vina wrote:
Amber G. wrote:P2

Problem from Junior Math Olympiad ...(For kids not yet in high-school)
Find (with proof) all n such that 2^n+12^n+2011^n is a perfect square..
(n is a natural number)


My answer is not an elegant "Proof", but rather a Paki like "Broof" , still ok I guess, since you said it is junior high level .

n =1 is the only possible answer.
Plugging in 0 for n, doesn't work. n = 1 works.

ie 2^1 + 12^1+2011^1 = 2025 = 45^2

Now
2^n + 12^n + 2011^n <= (2+12+2011)^n = 2025^n = (2011 + 14)^n

If we try expanding (2011 + 14)^n = (2011)^n + 14^n + positive terms

Re arranging a bit we have
2^n + 12^n + 2011^n - 2011^n - 14^n >= 0

ie 2^n + 12^n - 14^n >=0
{ AG Note:perhaps, as you say later...Flaw in the broof .. to get the above as one can't really get ">=" from "<=" in the previous inequalities..}

Now n = 1 is the only solution . For all n > 1 it is easy to see that LHS becomes negative (the function is decreasing with increasing n ) ie 14^n = (12+2)^n >= 12^n + 2^n + positive terms > 12^n + 2^n


Okay.. There is an error in the above workings. I just punched it off since I got busy. All the same, let me try to "Imbroove" the "Broof" and try to make it a dignified "Proof"

We have 2^n + 12^n +2011^n to be a square must be equal to (2011+ 2k)^n .. The LHS is odd, hence it should be an odd number that is raised to power n.

2^n + 12^n +2011^n = (2011 + 2k) ^n


{AG Note: RHS need not be a perfect power of n, just a perfect square..IOW n need not be equal to 2 }


Expanding the RHS as 2011^n + (2k)^n + (positive terms of lower order of n)

Since the LHS only powers of n, collectively the positive terms on the RHS of lower order must be zero for equality to hold

So rearranging, we have
2^n + 12^n +2011^n - 2011^n - (2k)^n = 0

or 2^n + 12^n - (2k)^n = 0

For natural numbers of n, putting n = 1, the min value of k = 7.

For values > 1, 2^n + 12^n - (2k)^n is less than 0 for all values of n (as discussed in the previous para). So k = 7 .

With k = 7, the only value that will satisfy the equation is n = 1 and that is the only solution for the problem!

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Re: BR Maths Corner-1

Postby Amber G. » 17 Jun 2011 22:56

Very good! Thanks.
vina wrote:
Amber G. wrote:Okay, here are two nice simple :) problems from a current national math Olympiads.. (Name of the country is omitted to make googling non-trivial :).. and the exam has passed so putting solution here is okay)

P1: (Problem is for high school kids)

Given, x, y, z are real positive numbers.. and
x+y+z = 3/670
and
1/x + 1/y + 1/z = 2010

Find, x, y, z
(All possible values, of course .. not just one or two...)


Multiplying the two we get
(x+y+z)(1/x + 1/y + 1/z) = 9
after simplifying and rearranging we got
(x/y + y/x) + (z/x + z/x) + (y/z + z/y) = 6
Now since we know that (a + 1/a) >=2 and since all of x, y, z > 0 , the only way this can be satisfied is when each of the terms within the brackets = 2 and from that it is easy to get that x=y=z is the only answer.

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Re: BR Maths Corner-1

Postby Amber G. » 17 Jun 2011 23:25

SwamyG wrote:http://www.artofproblemsolving.com/Store/index.php?

Anybody have any experiences in using the books from the "Art of Problem Solving" book store? My son enters the MathCounts competition, and I was wondering on the utility of these books.

I know the author Rusczek and others fairly well, and books are very good. (I have not used them, as for my kids I, never used books for teaching math.:). In particular, I have seen AOPS (Art of Problem Solving - two volumes) book and many like it a lot.

Nice program MathCounts (have been involved in it for quite some time). Hope your son enjoys it. (Did he compete this year?) If he is interested in math check out AMC (AIME/USAMO etc) competitions.. and USAMTS too.
(Here you get a set of 5 problems, you get about a month, and you can use web/books etc but you can not take others help)

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Re: BR Maths Corner-1

Postby SwamyG » 18 Jun 2011 02:09

^^^
Thanks. Yes, my son enjoys it and he has been competing the last two years. This year, his school made it to the State and lost badly there - courtesy his teacher. She thought they needed no practicing or coaching. Their team got creamed, they were 21st out of 28 teams. He is in the UMTYMP program.

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Re: BR Maths Corner-1

Postby vina » 18 Jun 2011 07:39

Amber G. wrote:Sirji - confused only...:) (See notes inside)

{ AG Note:perhaps, as you say later...Flaw in the broof .. to get the above as one can't really get ">=" from "<=" in the previous inequalities..}


Yes. Indeed that was the error in the "broof"

2^n + 12^n +2011^n = (2011 + 2k) ^n

{AG Note: RHS need not be a perfect power of n, just a perfect square..IOW n need not be equal to 2 }


Now in the "imbrooved" "broof" and hopefully it becomes a "proof", what I am essentially arguing is that for the LHS to be a whole power of n, the solution has to be of the order of the highest number raised to power n , for all n > 1 (n = 1 makes it a linear and not a higher power eqn). I just typed that in before hitting the sack last night and really didn't flesh it out.

So this power of 2 in the question is only a "trick" / "extra value" . Even if it were worded as a "whole power" , the answer would remain the same.
In my opinion, the key to finding the answer is that for any value grater than n =1 , the order has to be the same of n and the "solution" of n =1 is the exception and it just happens to be some convenient number raised to whatever power. You could set up a whole range of questions that way, but all have essentially the same answer,

For eg, instead of the question, if the question were,

11^n + 3^n + 2^n , find all values of n where it is perfect equation of order 4 , the answer is still n = 1 and the answer here is conveniently 2^4! .. Similarly 5^n + 2^n + 1 as a perfect cube has to be 2^3 and nothing else and for also stuff like 1 + 3^n + 23^n giving a perfect cube etc..

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Re: BR Maths Corner-1

Postby Amber G. » 18 Jun 2011 08:44

Vinaji ... unless you are :P ing ..I am lost.. (Does it makes sense to anyone else? :)
Anyway.. I don't understand why what you are saying has to be true ..

We have 9^n +40 ^n is a perfect square for n=1, but it is also true for n=2 Check it out.
Or 1^n+2^n+3^n+4^n+5^n+6^n+7^n+8^n is perfect square for n=1, it also true for n=3..

It may be hard to find such numbers.. but unless you show why, just saying that n=1 is the only solution, is no broof.

How about, n^2-19n+89? for what values it is a perfect square?

*****

The proof given by Nanjunamar is quite elegant.. Basically it says,
When n is even, LHS when divided by 3 leaves remainder 2 which is not possible to get a perfect square.
(All perfect squares are either divisible by 3 or leaves a remainder of 1 when divided by 3)
When n is odd and greater than 1, LHS when divided by 4 leaves a remainder of 3, which is not possible.
(All perfect square are either divisible by 4 or leaves a remainder of 1 when divided by 4)

Hence only odd number left to try is n=1.

(Notice that 12 is divisible by 3 and 4)

****

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Re: BR Maths Corner-1

Postby vina » 18 Jun 2011 12:47

Amber G. wrote:Vinaji ... unless you are :P ing ..I am lost.. (Does it makes sense to anyone else? :)
Anyway.. I don't understand why what you are saying has to be true ....

The proof given by Nanjunamar is quite elegant..


Indeed it is and quite intuitive as well. However, I think in the process of "broof" and "Proof" and hasty typing without putting pen on paper, I might have confused everyone. So let me try putting my thoughts in a more structured manner and lets see if it makes sense.

In the expression 2011^n + 12^n + 2^n , find all square values for natural n , Let us start by putting n = 1 .

2011 + 12+ 2 = 2025 = 45^2 , a square value. So the miniunum value the expression evaluates to is 2025 .

Now all other square answers if any have to be above 45 and be above 2025 , and since the expression is odd, let us assume for sake of ease of algebra it is of the form x^2 = (2011 + 2k) , with k >= 7.

Since the expression is of order n, final value too has to be of order n.

Assuming the that the expression you gave indeed evaluates to a square
2011^n + 12^n + 2^n = (2011 + 2k) ^n

What i showed earlier was that this cannot be true for any value other than n =1 , because for n >1, the term of the right will always be greater than the LHS because 12^n + 2^n < (2k)^n for all k >= 14! .

Proving n cannot be other than 1 is a sufficient condition for proving this problem. In fact that is what Nanjunamar did as well, in that way it is not different.

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Re: BR Maths Corner-1

Postby vina » 18 Jun 2011 13:11

Amber G. wrote:We have 9^n +40 ^n is a perfect square for n=1, but it is also true for n=2 Check it out.


If you notice the "broof" said that it is so because , 2^n + 12^n < 14^n for all n >1 and k >= 7

Now let us look at this case here. equivalently,

(40 + (2k+1))^n >= 40^n + 9^n .. For equality to hold, 40^n + (2k+1)^n = 40^n + 9^n , (2k+1)^n = 9^n which obviously holds for k >= 1 for n > 1 . (for k =4) So , that not true.

To be analogous to the problem you stated initially, this should be formulated as 2^n + 7^n + 40^n. This will have only one solution of n =1 .

So the "broof" (it is for a particular set where the relations are in the form that it reduces to) is not wrong I think.

1^n + 2^n + 3^n + 4^n + 5^n + 6^n + 7^n + 8^n is perfect square for n=1, it also true for n=3..


Let me try the "broof" . So for n =1 it sums to 36 ,so lets say that it is (8 +2k)^n again per convenience and k =14 when n=1

if we try 1^n + 2^n + 3^n + 4^n + 5^n + 6^n + 7^n + 8^n < 8^n + (2k)^n for all n > 1 and k>= 14, to be true

ie 1^n + 2^n + 3^n + 4^n + 5^n + 6^n + 7^n < 14^n = (2+3+4+5)^n , we see that this is not true , so yeah, there is more than one solution for this.

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Re: BR Maths Corner-1

Postby Amber G. » 18 Jun 2011 19:50

Since the expression is of order n, final value too has to be of order n.

What exactly does "order of n" means here. LHS is just a number, how do you determine the order of a number? :-?

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Re: BR Maths Corner-1

Postby Amber G. » 18 Jun 2011 20:57

Folks - Diophantine equations (see wiki, if not familiar with the term)could be interesting and could be very hard.

Famous (Fermat's last theorem) has been proved only recently.. (it says that
x^n+y^n = z^n has no (nontrivial, integer) solutions for n>2

There are infinite solutions for n=2, and it is easy to prove that it has no solutions for certain values of n (say n=4), but general proof is quite hard.

There is $100,000 prize, not to mention fame, (See:http://www.math.unt.edu/~mauldin/beal.html to either find a counter example (or prove it is impossible) for an equation of the kind
x^a+y^b = z^c (a,b,c >2 and certain restrictions on x,y,z - see the above link.

So, Vinaji ( :) ) and others...try to see if that method works...

Here are a few "simple" problems (I will give the answer in a few days to see if there are easier methods) Consider all variables are natural numbers in this discussion..

1: Are there solutions for x^3+y^3=z^2 .. if so how many finite or infinite?
2: How about x^3+y^3 = z^5 ?
3: I gave an example 1+2+...8 as a perfect square, can you find another one where
(1+2+3+....n) is a perfect square?

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Re: BR Maths Corner-1

Postby vina » 19 Jun 2011 12:31

Amber G. wrote:3: I gave an example 1+2+...8 as a perfect square, can you find another one where
(1+2+3+....n) is a perfect square?


The example you gave 1 + 2+... 8 is the only one which is a perfect square. There is no other perfect square for sigma n.

sigma n = n (n+1)/2 to be a square, n has to be of the form 2k^2 and
n+1 ie, 2k^2 + 1 has to be a square as well, say as y^2

Now let us start plugging in values.

0 and 1 dont work. For k = 2, it works . Now all that is left to "broove" that for k>2, no solution exists.

since 2k^2+1 = y^2 = (2k+x)^2 ... for k >2 , where x is a positive number

ie 2k^2 + 1 = 4k^2 + 4kx + x^2

or 2k^2 + 4kx + x^2 -1 = 0 . This is impossible because for k >2, the term 2k^2 + 4kx + x^2 is always > 1 . So the "broof" sort of works.
Last edited by vina on 19 Jun 2011 13:41, edited 1 time in total.

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Re: BR Maths Corner-1

Postby vina » 19 Jun 2011 13:41

Amber G. wrote:1: Are there solutions for x^3+y^3=z^2 .. if so how many finite or infinite ?


Hmm. I never include 0 in natural numbers,but I looked up wiki and it seems that there is a "controversy" . It is infinite if you include 0 , and finite if you don't with 1, 2 , z=3 the only values.. Though this is the right answer, let me attempt a proof (hopefully not a "broof") when I have the time pls. The "broof" I am positive will work

As for the other question x^3 + y^3 = z^5 there is no solution. Isn't that a corollary of the Fermat's theorem x^n + y^n = z^n . , if you don't include 0 as a natural number . If you do, there are infinite answers

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Re: BR Maths Corner-1

Postby Amber G. » 19 Jun 2011 20:40

vina wrote:
Amber G. wrote:3: I gave an example 1+2+...8 as a perfect square, can you find another one where
(1+2+3+....n) is a perfect square?


The example you gave 1 + 2+... 8 is the only one which is a perfect square. There is no other perfect square for sigma n.

sigma n = n (n+1)/2 to be a square, n has to be of the form 2k^2 and
n+1 ie, 2k^2 + 1 has to be a square as well, say as y^2

Now let us start plugging in values.
<rest of the p(b)roof omitted.. see above post for details >



Are you sure? How about (1+2+...49) which is 1225 = 35^2...
(or looking at it as 49*(50/2) = 49*25)
Or even (1+2+3....1681) which is 1189^2
:)
Last edited by Amber G. on 19 Jun 2011 23:20, edited 1 time in total.


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