## BR Maths Corner-1

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Amber G.
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### Re: BR Maths Corner-1

Schmidt wrote:

We can use a^3 + b^3 >= a^2b + ab^2
...

Excellent. This concept (Rearrangement Inequality #note 1), is, in my opinion, intuitively more obvious and powerful than AM/GM type concept. (This is what I was thinking when I made the problem)

If there are \$100 bills and \$1 bills, and you have to pick 10 bills of one kind, and 1 bill of the other, even a child will pick 10 bills of \$100 and 1 of \$1 rather than the other way. (In general a*a^2 + b* b^2 >= a * b^2 + b* a^2 === same concept)

So rewriting what you wrote:
given abc=1
a^3+b^3 >= ab (a+b) = (a+b)/c ==> a^3+b^3+1 >= (a+b+c)/c ==> 1/(a^3+b^3+1) <= c/(a+b+c)
Rest is simple.
(Original problem a^5+b^5+ab reduces the same way by noticing a^5+b^5 >= a^2*b^2(a+b))
***
Unfortunately I have not seen this (Rearrangement Inequality) covered in a typical math class. Though it is quite powerful - one can easily derive AM/GM or power-mean or Cauchy–Schwarz Inequality from this.

Intuitively obvious, this inequality is: (/taking lazy way out and not to write a longer post / --> Just look it up in wiki ) .
One more powerful part is, unlike AM/GM the values here don't have to be positive.
Last edited by Amber G. on 25 Jun 2020 00:51, edited 1 time in total.

Vayutuvan
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### Re: BR Maths Corner-1

Wow. I really like that Rearrangement Inequality. I did not come across it before. The one I know best is Cauchy-Schwartz Inequality and AM > GM.

The problem you have given above was posed in the Facebook group called Inequalities. One proof made the same mistake as above. I pointed out the mistake, but whatever I tried, I couldn't fix it.

Amber G.
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### Re: BR Maths Corner-1

^^^Thanks, glad it was enjoyable. Yes, this inequality is not as well known, but in my coaching this is one concept I *always* cover (along with Power-Mean, AM/GM/HM and Cauchy-Schwartz - and perhaps Holder, Jensen etc.- my approach is more of intuitive understanding rather than formal treatment).

With this, for example a*a+b*b >= a*b +b*a is nothing but AM/GM.

Also it is *very* intuitive - If one has to pick, say (a,b,c) bills from a stack of \$100, \$10, \$5 etc bills, .. and given a>=b>c , best strategy is to pick a \$100 bills, b \$10 bills, c \$5 bills etc..).

Another way to see this is intuitively obvious, as any physicist or a kids can see, is to think of bunch of kids (weights - x_1, x_2, x_3..) sitting on a sew-saw at distances (y_1, y_2,. y_3...)from the fulcrum. The greatest turning moment around the fulcrum will be obtained when the heaviest kid sits the farthest away from the fulcrum .. and so on. This is nothing but the arrangement inequality.
---
Both US and India's olympiads (and JEE type exams) and International Math Olympiads often have such problems of inequalities, so learning these tools are good for exams too .
---

sudarshan
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### Re: BR Maths Corner-1

Amber G. wrote:Formal solution, as I posted before is:

x = ( (3+2 √ 2)^n - (3 - 2 √ 2)^n) / (4 √ 2)
y = ((3+2 √ 2) ^n + (3 - 2 √ 2) ^n - 2)/ 4

Answers are given by choosing n=0,1,2,3,....

Surprising the "practical" solution is *VERY* easy as 3+2 √ 2 ≈ 5.82842712475 all you have to do is to find (5.82842712475^n) .. for example
y=(5.82842712475^n - 2)/4 (Just take the integer part only .)

OK, but how does one arrive at this solution, algebraically?

---
In some of the books, I have seen the formula is credited to Euler, but of course, this was well known to Brahmgupta and *many* others for thousands of years.

The method, as you can see, is fairly well known, easy to prove (extremely easy to prove). Interesting this theme does appear many times in competitive exams.
(Similar problem I asked here in brf before - find a right-angle triangle with integer sides (a,b,c) such that b=a+1. (Examples (3,4,5))
-------
Sudarshanji - Just curious, did you find any other solution to my first problem - how high up did you (or your computer program) go? TIA.

The first problem meaning - the sum of integers up to y being equal to the sum of squares up to x? I don't have a computer program, was doing it by hand in Excel, so only got the first two <x,y> pairs. Even the easier problem was set up in Excel.

sudarshan
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### Re: BR Maths Corner-1

Amber G. wrote:Okay - here is a simple problem (as easy as 1,2,3 ) I posed in a group consisting of mainly Google Engineer's

Find three positive rational numbers such that their sum is 6 and the product is 6.

One answer is 1,2,3 as 1+2+3 = 1*2*3 = 6
(Only fractions - rational numbers which are positives - are allowed.
(Let us see if brfites visiting this forum can find the answer(s) before the other group... computers /books etc are allowed .. googling is discouraged but I can't stop it )

(Request: Please For this problem (only) _ No invoking great mathematicians and writing about fantastic theorems ... Just the answer(s) please.. if one answer appears find a different answer - Imagine your audience just know how to add and multiply ordinary fractions! I want to see how many answers we get here in brf ...)

So this problem is:

abc=6;-------a+b+c=6

For a given a:

b+c=6-a;-------b(6-a-b)=6/a

Working with that last equation:

b^2-(6-a)*b+6/a=0

b=0.5*( (6-a) +/- sqrt( (6-a)^2 - 24/a ) )

For rational b, the discriminant has to be positive:

(6-a)^2 >= 24/a

(6-a)^2 is a parabola with a minimum at a=6; 24/a is a rectangular hyperbola with a discontinuity at a=0. We want all a where the hyperbola is below the parabola.

The hyperbola will be below the parabola for all a<0. There are also three points where the parabola and hyperbola have the same value (since this is a cubic equation). Too lazy to find the exact values of a, but roughly:

Root 1 (call it R1): a is between 0.9 and 1
R2: a is between 3 and 4
R3: a is between 7 and 8

For a<0, the hyperbola is of course below the parabola, but a can't be negative, so this case is out.

For 0<=a<R1, the hyperbola is above the parabola, hence this case is out.

For R1<=a<R2, the hyperbola is below the parabola (which is what we want).

For R2<=a<R3, the hyperbola is above the parabola, hence this case is out.

For R3<=a<Inf, the hyperbola is below the parabola, and a is also positive, but b and c will both turn out negative, hence this case is out.

So any value of a between R1 and R2 will work, except for the condition that a, b, and c all have to be rational. Not sure how to enforce this condition.

For a given a, it is easy to see that b will be one solution to the quadratic, and c will be the other solution - i.e.:

b= 0.5 * ( (6-a) - sqrt( (6-a)^2 - 24/a) )

c= 0.5 * ( (6-a) + sqrt( (6-a)^2 - 24/a) )

A further point - a=1, b=2, c=3 is a solution, but this is technically the same solution as a=2, b=1, c=3, or a=2, b=3, c=1, etc. So a further condition could be enforced, that a<=b<=c. This could modify the range of a (restrict it from the current R1<=a<=R2).

sudarshan
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### Re: BR Maths Corner-1

<a, b, c>=<49/15, 25/21, 54/35> (Rearrange in ascending order if so desired).

Check it out: a+b+c=6, abc=6

Obtained by a little logic and then trial and error, will elaborate in a while.

Amber G.
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### Re: BR Maths Corner-1

<a, b, c>=<49/15, 25/21, 54/35> (Rearrange in ascending order if so desired).

Check it out: a+b+c=6, abc=6

Obtained by a little logic and then trial and error, will elaborate in a while.

Great! Many (most) find (8, -3/2, -1/2) -- numbers which are not all positive, so I put the condition "positive" part.
Let us see if more such numbers are found! Amber G.
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### Re: BR Maths Corner-1

This is a Gem! For Ramanujan's 100th year - a webinar by none other than prof Professor Bruce Berndt. Save the link watch when you can.
Lot of Math, Photographs stories etc..
https://youtu.be/cAjMMvQIY9o

There was just today

A few days ago there was a presentation by Prof Ribet I will post the youtube or video link if there is some interest here. VERY nice lecture about Fermat's Last Theorem. (I do have PP slides of the lecture - will try to share it here). (The background needed to understand is graduate level math but need not be expert -- It is one of the most understandable lectures I have seen for FLT)

(BTW - with those two lectures .. the last few problems I posted would be child's play )

sudarshan
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### Re: BR Maths Corner-1

Amber G. wrote:Great! Many (most) find (8, -3/2, -1/2) -- numbers which are not all positive, so I put the condition "positive" part.
Let us see if more such numbers are found! I tried:

Let a=x/y where x and y are both integers

b = 0.5 * ( (6-a) - sqrt( (6-a)^2 - 24/a ) )
c = 0.5 * ( (6-a) + sqrt( (6-a)^2 - 24/a ) )

b = ( 1 / ( 2 * y * sqrt(x) ) ) * ( sqrt(x) * (6*y-x) - sqrt( x*(6*y-x)^2 - 24*y^3 ) )
c = ( 1 / ( 2 * y * sqrt(x) ) ) * ( sqrt(x) * (6*y-x) + sqrt( x*(6*y-x)^2 - 24*y^3 ) )

For both numerator and denominator for b and c to be integers, one way could be that x is a perfect square (say k^2), with the additional condition that k * (6*y-x)^2 - 24*y^3 has to be a perfect square. Trying combinations of k and y, I kept getting <1,2,3> (various permutations of that) plus <8,-1.5,-0.5>, plus another solution which had b & c both being negative. But then k=7, y=15 yielded the solution above: <49/15, 25/21, 54/35>.

But of course, this is not the only way to get rational b & c.

Amber G.
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### Re: BR Maths Corner-1

^^^ Very good!
How about trying some more numbers .. like k=143, and y=8023 or something like that. (Just curious, what other set you found where b,c were negative (other than -3/2, -1/2))..

sudarshan
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### Re: BR Maths Corner-1

Amber G. wrote:^^^ Very good!
How about trying some more numbers .. like k=143, and y=8023 or something like that. (Just curious, what other set you found where b,c were negative (other than -3/2, -1/2))..

k=143, y=8023 worked: <a, b, c> (ascending order) = <15123/16159, 25538/10153, 20449/8023>.

Going much higher than that runs into issues with even .NET long (int64) types.

The other set with negatives that I found was:

<a, b, c> (ascending order) = <-361/68, -32/323, 867/76>.

Vayutuvan
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### Re: BR Maths Corner-1

Amber G. wrote:This is a Gem! For Ramanujan's 100th year - a webinar by none other than prof Professor Bruce Berndt. Save the link watch when you can.
...
(BTW - with those two lectures .. the last few problems I posted would be child's play )

That reminds me, Prof. Berndt is retiring. He has several books and a stack of papers he ever published. I am supposed to go and root through the material and take whatever I can. He has biographies of several Indian mathematicians as well as the book "Vedic Mathematics". Our common friend, who is a vegan and yoga practitioner asked him about the book. Prof. Berndt answered thusly: "Hindus had a very advanced mathematical system thousands of years back".

I consider myself lucky to have known him.

Amber G.
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### Re: BR Maths Corner-1

^^^Thanks. I highly recommend the recent talk I posted, it is very good. Prof Bruce Berndt, along with others - many pieces written/edited/compiled with Freeman Dyson and him about Ramanjuan's work are very good (my personal favorites). UVA's Ken Ono, recently has given some good popular lectures about Ramanujan's work too.

Say Hi to prof. (last few problems in brf may amuse him - as they are inspired/compiled from Ramnujan's very popular work. )

For those who may not know - Prof Bruce Berndt's multi-volume books about "Ramanjuan's Papers/notebooks" is *very* good, it gives proofs of many conjunctures of Ramnujans and comments on insights of many of his concepts - It is a *very* good reference for anyone interested in this aspect. (There was an effort - may be still going on - to compile an wikipedia/encyclopedia type writing of all of Ramanujan's work).

Amber G.
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### Re: BR Maths Corner-1

sudarshan wrote:
Amber G. wrote:^^^ Very good!
How about trying some more numbers .. like k=143, and y=8023 or something like that. (Just curious, what other set you found where b,c were negative (other than -3/2, -1/2))..

k=143, y=8023 worked: <a, b, c> (ascending order) = <15123/16159, 25538/10153, 20449/8023>.

Going much higher than that runs into issues with even .NET long (int64) types.

....

As some may have guessed, there is a fairly simple method to attack these kind of problems.. one can get generate these numbers, as many as you want, fairly easily - once the method is known - the fun part is - the numbers go big quite fast so without theory brute force calculation can only produce a few numbers. (There are infinite set of such numbers)

Some math theory here is quite interesting, and gets interesting and complicated. When I have some time (and if there is interest) I may put some comments here, but standard math courses in number-theory cover this.

Amber G.
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### Re: BR Maths Corner-1

Another interesting old problem, which just have been solved (lot of buzz in math world) is "square in a loop" problem. Basically this is an old problem, very easy to describe:

Draw a close loop. (Imagine you started from point A, took a trip, went to many different places and came back to point A): The conjecture was that one can *always* find 4 points on that closed loop / path such that these four points will make a square.

(People are claiming that it has been proven - I have not looked at the proof but was interested in that problem over the years).

Vayutuvan
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### Re: BR Maths Corner-1

Amber G. wrote:As some may have guessed, there is a fairly simple method to attack these kind of problems.. one can get generate these numbers, as many as you want, fairly easily - once the method is known - the fun part is - the numbers go big quite fast so without theory brute force calculation can only produce a few numbers. (There are infinite set of such numbers)

I am thinking more on the lines of start with (1,2,3), (1,3,2), (2,1,3), (2,3,1), (3,1,2), (3,2,1) and search intheir vicinity.

Let us assume that all three are equal to 2. Then their product is larger than 6. If one is close to 6, the other two have two are bounded fro below. If two are equal say close to 3, then the other is bounded from below by 2/3.

Let us assume that the three rational numbers are a/x, b/y, and c/z and also (a,x), (b,y), and (c,z) are pairs of relative primes. We need x, y, and z to be greater than 1 as well as a > x, b> y, c > z.

One equation we can get is cyc+(1/((b/y)(c/z)) = 1

where cyc+ denotes the cyclic addition.

...
?

Amber G.
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### Re: BR Maths Corner-1

Ronald Lewis Graham - One of the principal architects of the rapid development worldwide of discrete mathematics in recent years died today.

Vayutuvan
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### Re: BR Maths Corner-1

That is sad. Great man. I have the book "Concrete Mathematics" Knuth co-authored with Graham and Patashnik. I read large parts of it. but need to attmpt problems.
Last edited by Vayutuvan on 08 Jul 2020 23:00, edited 1 time in total.

Vayutuvan
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### Re: BR Maths Corner-1

https://writings.stephenwolfram.com/2016/04/who-was-ramanujan/

I highly recommend this blog article by Wolfram. He has great insights. I never met Wolfram even I passWolfram Corp. HQ on my way to work and back.

Amber G.
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### Re: BR Maths Corner-1

Vayutuvan wrote:That is sad. Great man. I have the book "Concrete Mathematics" Knuth co-authored with Graham and Patashnik.

He was also known for fun math - like magical tricks, and kind of problems yours truly puts in this thread. I think he contributed a lot for Martin Gardener's classic column (Mathematica Games which ran for decades and was very popular).. He will be missed by many.
Here is a picture from a few years ago: (With his Magical Mathematics Book) Amber G.
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### Re: BR Maths Corner-1

Speaking of Ramanujan .. These slides from a recent lecture by Prof Henri Darmon puts my April 1, post from some time ago in a whole new perspective.. . (BTW Darmon's lecture, as usual was excellent - you can find the slides and video which will be posted fairly soon on the university's cloud platform so just search for it)

Old post: Here:
(This was an April Fool Joke, BTW ..)

Amber G. wrote:On this day let me bring you an exciting story....

Anyway I got this email from a math friend who was working on some old Ramajnujan problem regarding Heegner theorem. She found an interesting result. The result is not published yet so you are reading it first hand here.

Let me present it here - she found that amazingly

(ln(12^3(231^2-1)^3+744)/pi)^2 = 163

Here ln = natural logarithm (that is base e)
Pi = 3.14159265....
etc..

(If you don't believe it, just use a calculator / google to check it out)

Of course, the result is rather surprising. Pi is an irrational number. And natural logarithms of even ordinary numbers are almost always irrational so finding an integer result of this rather complicated expression is VERY remarkable. Naturally it has baffled all those mathematicians who were on this email list. Once the result is published I think it will cause a major revolution in number theory.

I am sure, my math friends know that 163 is largest Heegner number - that is this is the largest number where quadratic imaginary number fields admit unique factorisation in their ring of integers.

Slides from the recent lecture, which many may find interesting or at least be amused by it -- and impressed by genius of Ramanujan. <... and.. > (For more, see Ramanujan's Constant say here: https://mathworld.wolfram.com/RamanujanConstant.html#:~:text=Numbers%20such%20as%20the%20Ramanujan,this%20sort%20of%20near%2Didentity.

Vayutuvan
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### Re: BR Maths Corner-1

One quick question @Amber G. Would Prof. Bruce Berendt recognize you by Amber G.?

Amber G.
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### Re: BR Maths Corner-1

^^^Vayutuvanji No. (But even if someone comes to know about more private information than I have shared here, please do not post. Thanks.
(He will certainly be familiar with the type of last few problems I posted and it will be *very* easy to give the answers or provide some very easy way to solve, so it will be very interesting to see his reaction)

Amber G.
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### Re: BR Maths Corner-1

Another, *very* interesting popular lecture from Prof Maryna Viazovska ( Invited lectures series shared by UVA on zoom.
How one can pack spheres in n-dimensional space with lot of interesting number-theory results and it's relation to physics.
(I think many here will like it)

https://virginia.zoom.us/rec/play/tcJ4dumtrjg3SYaR5QSDUf5wW420ev-s1iEZ_6ZcnkewUSEFOlX3YbVDZbd2uArG2Ba1c2_kp8JPdYOk?fbclid=IwAR1b1oEAZ_nXvkGhiJYTjGq6w-rMIfU3lLzlRheXuOhQGXtUSqbPZs1GjeU

Vayutuvan
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### Re: BR Maths Corner-1

Amber G. ji, no problem. I won't.

Does the "add/multiply to 6 problem" has anything to do with continued fraction expansion of sqrt(19)?

√19 = [4;2,1,3,1,2,8,2,1,3,1,2,8,...] (sequence A010124 in the OEIS). The pattern repeats indefinitely with a period of 6.

Amber G.
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### Re: BR Maths Corner-1

^^^ No, the continued fraction is related to Diophantine equations of second degree. Integer solutions for first or second degree equations are relatively easy and there are many standard methods.
For Cubic equations, the calculations are a little more involved.

For example, the sum/product = 6 equation is basically standard cubic equation y^2 = x^3 - 9x + 9 ;

Every integer solution x, can give three numbers (a,b,c) where a = 6/(3-x), b = (6 - 3x + y )/(3-x) and c= (6 - 3x - y)/(3-x)
(Check out, that a+b+c =6 and abc =6) so every rational x, will also give rational (a,b,c). (x != 3).
(In addition if x<2 all three values will be positive)
***
Easy to see x=0, (y^=9), x=1 (y^2=1), gives rational solutions - As said before there are standard methods to find out, if there are finite or infinite values in integer (or rational) domain and if there are infinite values, one can find as many as one wishes .. again standard methods (unfortunately not taught that much in ordinary math classes -

***
Some math is not that hard (that's why I posted here in brf, but theory and some concepts here - like modular functions etc are quite important in modern math and do get quite involved. BTW Fermat's Last Theorem depends in reducing x^n+y^n=z^n into a form very similar to
above equation (y^2 = x^3+....) and then proving it has no solutions in rational domain)... basically proving by contradiction that if FLT had one solution then t can write it the form of above equation etc... (Okay I am simplifying 600 page proof into a post . ).

****

Vayutuvan
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### Re: BR Maths Corner-1

Amber G. ji, thanks for the simple explanation. While trying to solve this simple problem, I went through LeVeque's book very quickly after Wikipedia pages on rationals, diophantine approximations, and finally just now through the fourth chapter of Graham-Knuth-Patashnik's Concrete Mathematics. They have a pretty good amount of stuff on mod function, Euler's Totient function, Mobius function mu, and a very very nice exposition of Stern-Brocot tree.

That said, somehow I never liked Concrete Mathematics style of exposition. It is very conversational. I have the same problem with Strang's two books - Linear Alegbra and Introduction to Applied Math. The latter is a little better though since it touches on many Applied Math. areas.

Amber G.
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### Re: BR Maths Corner-1

Sudarshanji, Vayutuvanji et all - Okay, I am putting solution to my own problem - Posting the original problem in quotes for convenience, but there is quite a lot of discussion above.
Amber G. wrote:Okay - here is a simple problem (as easy as 1,2,3 ) I posed in a group consisting of mainly Google Engineer's

Find three positive rational numbers such that their sum is 6 and the product is 6.

One answer is 1,2,3 as 1+2+3 = 1*2*3 = 6
(Only fractions - rational numbers which are positives - are allowed.
(Let us see if brfites visiting this forum can find the answer(s) before the other group... computers /books etc are allowed .. googling is discouraged but I can't stop it )

We got some answers, (by Sudarshanji - see those posts) mostly I think, by trials . This type of problem is less common to be asked in ordinary exams and may be a little harder. Of course there are many resources where one can get more mathematical details and theory but few comments, some may find interesting.

- There are infinite solutions to this problem.. and one can systematically find as many as one want with simple algorithm.
Here is one way - where I just give the method. (As they say this method comes in a dream where Namagiri appears )
>>>

Now find the new value of x and y (call it x_1 and y_1) by the method given just below (Method #1 - recursion)
Using these new values of x_1 and y_1, use the same method to find new values of x and y (call it x_2, y_2) .
This way you can get as many values of x and y as you want..
(

for each value, the answer three values would be
6/(3-x) , (6-3x+y)/(3-x) , (6-3x-y)/(3-x)

Since x and y are rational, all three values above will be rational, and you can verify that sum and product of the above three numbers would be 6.
(Sum is easy to see that it is 6, the product, will also be 6, if you follow the method given below to find right x and y)

****
Method #1
calculate m = (17 - y_0)/(7- x_0)
(new) x = m^2 - 7 - x_0
(new) y = m (7- x) - 17
***
That's all, it's that simple.
For example, first set of (x,y) = 1,1
This gives m = (17-1)/(7-1) = 8/3
so new x = 64/9 - 7 - 1 = -8/9
and new y = 109/27

Now when x=1, y=1, (first value) gives the three values 6/2 , 4/2, 2/2 which is (3,2,1)
and the second set gives the values given above.
(6/(3+8/9); (6+8/3+109/27)/(3+8/9) ; (6+8/3+109/27)/(3+8/9)
which is 54/35 ; 49/15; 25/21
Now you can start with (-8/9 and 109/27 as x,y value and get a new set of values..
And repeat...

Amber G.
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### Re: BR Maths Corner-1

^^@sudarshanji - You may like to check the values you have gotten are same given by above method...

sudarshan
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### Re: BR Maths Corner-1

I iterated the way you said. So the next step after x=-8/9 and y=109/27, gives:

a=15123/16159; b=20449/8023; c=25538/10153

I verified that the sum and product are both 6 (this is also your earlier suggested solution with k=143, y=8023).

The next step after that gives numerator and denominator terms for a, b, and c, which are all in 15 or 16 digits (*after* simplification). But roughly:

a=1.516598646; b=1.207775855; c=3.275625499 (approximations of the rational fractions)

Again, abc=6, and a+b+c=6.

After that it's beyond .NET int64 again (at least, Excel blows up).

Maybe I'm doing something wrong, maybe it shouldn't blow up that fast?

So if I want to know why this particular iterative procedure works, do I need to read a 600 page proof? .

Amber G.
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### Re: BR Maths Corner-1

^^^Thx, for confirming. Some parts - (eg recursive relationship) are not hard and I have been able to explain/teach it to UG/High schools - some techniques are quite useful now in other subjects like encryptions etc as some process here can go in one direction quite easy but the other direction is not very easy.

Yes the numerators/denominators grow big *very* fast, and although one can get as many answers, in practice to get exact numbers is lot of work even with computers.

But *all* such numbers will appear in the above list, these numbers form a group (mathematical group theory) and have nice properties -

Anyway, iPhone encryption is based on an equation very similar to some problems asked here ... .. (Easy to go in one direction and get numbers but *very* difficult to go in other direction)

Vayutuvan
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### Re: BR Maths Corner-1

sudarshan wrote:ISo if I want to know why this particular iterative procedure works, do I need to read a 600 page proof? .

The published proof is about half as long Amber G.
BRF Oldie
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Location: Ohio, USA

### Re: BR Maths Corner-1

Salute to Shri C.S. Seshadri, who just passed away.

The passing of Professor C. S. Seshadri is a great loss to mathematics, science and teaching. He was among those who built the TIFR School of Mathematics and the Chennai Mathematical Institute.

This is the loss of an intellectual giant whose feet were firmly planted in the vibrant and welcoming institute he built, students from all over India come.

His work on projective modules over polynomial rings, geometric invariant theory, moduli theory, vector bundles on curves, the Narasimhan-Seshadri theorem, parabolic bundles, standard monomial theory, and the geometry of Schubert varieties is note-worthy.

Amber G.
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### Re: BR Maths Corner-1

sudarshan wrote:I iterated the way you said. So the next step after x=-8/9 and y=109/27, gives:

a=15123/16159; b=20449/8023; c=25538/10153

I verified that the sum and product are both 6 (this is also your earlier suggested solution with k=143, y=8023).

The next step after that gives numerator and denominator terms for a, b, and c, which are all in 15 or 16 digits (*after* simplification). But roughly:

a=1.516598646; b=1.207775855; c=3.275625499 (approximations of the rational fractions)

Again, abc=6, and a+b+c=6.

After that it's beyond .NET int64 again (at least, Excel blows up).

Maybe I'm doing something wrong, maybe it shouldn't blow up that fast?

So if I want to know why this particular iterative procedure works, do I need to read a 600 page proof? .

Btw, if one uses Mathematica, or Maple or even Rexx or Fortran (with many digits) you need not run into .NET int64 issues..
For the record I will just state here the exact result..::) ..

(29179626901489 / 29415245790105, 47083791468169 / 23156698179195, 110261725903350 / 37065988023371)  Vayutuvan
BRF Oldie
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### Re: BR Maths Corner-1

You know Rexx? Nice language. Lisp or Scheme also is capable of doing large numbers out of the box. Probably can be done in emacs with Elisp.

sudarshan
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### Re: BR Maths Corner-1

Amber ji, does that feature in Mathematica/.../... only work with integer math, or also with floating point?

I tried something similar in C myself a while ago. Basically a limitless int class, storing data as strings, and implementing basic operations like ASMD the same way one would do those on paper, i.e., one digit at a time with carryover. The string data base allows the numbers to get really long, limited only by RAM or disk space. There are also class libraries available online, which implement floating point numbers with arbitrary precision. I don't know what their base data type is, probably string or arrays of characters (with each character being limited to digits) or arrays of short ints (each short being limited to digits, or maybe three digits for efficiency) or something like that.

vsunder
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Location: Ulan Bator, Mongolia

### Re: BR Maths Corner-1

This paper posted on Arxiv on Monday solves a 30 year old problem on the nodal sets of eigenfunctions.

https://arxiv.org/pdf/2008.00677.pdf

A nodal set is the set where the eigenfunction is not zero. A cheap summary of the result is Nodal sets cannot be thin and narrow. A first version of this result was published by the Fields medallist and Princeton professor

https://en.wikipedia.org/wiki/Charles_Fefferman
in 1989. The results all belong to a set of conjectures by the Fields medallist and Harvard professor ST Yau:(read the intro to our paper for details and history)

https://en.wikipedia.org/wiki/Shing-Tung_Yau

Soon after in 1991 I had substantially improved these works along with my recently departed colleague Ben Muckenhoupt:

The key point in my paper with Ben was the subtle covering lemma 3. However the optimal result was nowhere in sight. Now this new paper gets the optimal result and finishes the problem.

A word on my collaborator AL on this new paper. He is a star and I had real pleasure working with him over 2 years on this problem. I consider him to be the best young person in Geometry and Analysis today and even stronger than Figalli who won the last Fields medal. AL has won several prizes already, the Clay Math. prize, the prize of the European mathematical Society and the Grand prix de Salem

https://en.wikipedia.org/wiki/Aleksandr ... hematician)

He is currently asst. prof at Princeton and he will go very far. As a matter of interest the Salem Prize in honor of French mathematician Raphael Salem was awarded in 1969 in its second year to my thesis advisor Richard Hunt for his proof of the famous Lusin conjecture after 60 years of the convergence for Fourier series and in 1984 to my friend and collaborator and current president of the International Mathematical Union

https://en.wikipedia.org/wiki/Carlos_Kenig

Out of the papers I wrote with this guy ^^^ I particularly want to single out our joint work on the free boundary of composite materials which was very hard, took us a long time and eventually we cracked it. Here are the two works:

https://sites.math.rutgers.edu/~chanillo/chankenig.pdf

https://sites.math.rutgers.edu/~chanillo/0804.pdf

Raphael Salem

https://en.wikipedia.org/wiki/Rapha%C3%ABl_Salem

Vayutuvan
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### Re: BR Maths Corner-1

sudarshan wrote:Amber ji, does that feature in Mathematica/.../... only work with integer math, or also with floating point.

Best bet is to use rational field type along with large number arithmetic. There are several packages including probably in Boost as well. One version of standard Lisp I had on my IBM DOS machine back in 1987-88 timeframe, had bignums (integer field only) built-in. I don't remember now where I got that Lisp interpreter from. I bet you can get a good scheme/lisp interpreter from GNU.

Alternatively, go for a free symbolic algebra system like Macsyma. The FLOSS version - Maxima - is here.

http://maxima.sourceforge.net/

Mathematica is top expensive while reasonably good alternatives (not as flashy) exist in FLOSS.

Added later: Looks like Maxima does not have arbitrary precision arithmetic. I am so darn sure I had that on my common lisp interpreter from GNU/MIT.

Well, better to go for a more modern up-and-coming language then. Here is a link to Julia.

https://docs.julialang.org/en/v1/manual/integers-and-floating-point-numbers/#Arbitrary-Precision-Arithmetic-1

If you really want to experiment with Number Theory, then Sage is really good in that respect. They have a Sage server/cluster free for use. Everything is in Python though.

https://sagecell.sagemath.org/

Commercial sage (with a free personal account with limits on RAM and disk space) is here
https://cocalc.com/

sudarshan
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### Re: BR Maths Corner-1

Vayutuvan wrote:Best bet is to use rational field type along with large number arithmetic. There are several packages including probably in Boost as well. One version of standard Lisp I had on my IBM DOS machine back in 1987-88 timeframe, had bignums (integer field only) built-in. I don't remember now where I got that Lisp interpreter from. I bet you can get a good scheme/lisp interpreter from GNU.

...

Commercial sage (with a free personal account with limits on RAM and disk space) is here
https://cocalc.com/

Thanks! I don't have a real need for it, except as a curiosity or to handle problems that come up on BRF. Julia sounds interesting, of the choices you listed. Will take a look.

Amber G.
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### Re: BR Maths Corner-1

- Yes I use Rexx. It has unlimited precision (just define as numerical digits (3000) and all calculations will be carried out to 30000 digits). I still use Fortran, (along with a few other languages) which also has subroutines etc.. for increased precision. (I wrote one my self in 1966 in IITK where I was having fun calculation Pi to a large number of decimal places on our Institute's IBM 7044).
(Rexx is still available on all platforms - just do google - many versions are free and quite good - there are even online version where you can do calculations on the cloud)

- Both Mathematica, and Maple (and virtually all other packages) have this type of capability. I have seen Basic (in 80's, modified to work with many digits) available from universities etc. .. Just do google one can get many such programs. Choose one. Favorite package depends on the chooser.

(They cost money but connection with a university etc can make it easier to get -- there are plenty free online versions too).

****
Meanwhile I happened to read some interesting history - an old Sanskrit text where a problem was asked almost 1500 years ago, given with solutions by Brahamgupta. The same problem was asked by Fermat about 1000 years later and he has to wait for the solution by Euler (or some other person for about 100 years --- Brhamgupta's that particular text was translated in English (from an Arabic translation) in 1700 or 1800..
Anyway that was interesting. (I do have a copy of an old sanskrit text where that method is discussed - quite fascinating read.)

Here is that problem, it is very famous: (If you want to try out your computer skill)..

Take a number (integer say n)
square it (n^2)
Multiply it by 61 (61n^2)