Thoughts On The Abc Conjecture

[phillip1882]

i found an interesting algorithm for producing potential liars to the ABC conjecture.

the algorithm is as follows.

let a,b,c be a primitive Pythagorean triple, with b even.

then both (b-a)^2,4*b*a,(b+a)^2 and (b-c)^2,4*b*a,(b+c)^2 are possible liars.

that is where log C/log rad(A*B*C) > 1.

now, there's no guarantee of course. the larger the pythagorean triple, the more factorable the terms have to be.

 

but it seems to me that in order to prove the ABC conjecture, you would have to prove that there are finitely many pythagorean triples that lead to good liars; a daunting task.

[Turtle]

i found an interesting algorithm for producing potential liars to the ABC conjecture.

the algorithm is as follows.

let a,b,c be a primitive Pythagorean triple, with b even.

then both (b-a)^2,4*b*a,(b+a)^2 and (b-c)^2,4*b*a,(b+c)^2 are possible liars.

that is where log C/log rad(A*B*C) > 1.

now, there's no guarantee of course. the larger the pythagorean triple, the more factorable the terms have to be.

 

but it seems to me that in order to prove the ABC conjecture, you would have to prove that there are finitely many pythagorean triples that lead to good liars; a daunting task.

 

 

Crap!! Just lost a post to a LOS with 20 minutes invested.

 

The short of it, there's a distrubted computing project called ABC @ Home that's doing a one-by-one search & now headed to 10²⁰. The idea is that with a complete list of "liars" [your term I presume for d's > c's??] on some interval from 1 it might be possible to see the daunt & call it out.

Edited June 21, 2013 by Turtle

[phillip1882]

just for the "fun" of it, i wrote a program to produce a couple hundred Pythagorean triples that lead to abc conjecture "liars" yes turtle, my term for cases where its "not true".

abcdata.txt

here's the code itself.

import math

def gcd(a,b ):
  if a == 0:
     return b
  if b == 0:
     return a
  if a < b:
     return gcd(a,b%a)
  else:
     return gcd(a%b,b )

def rad(n):
  i = 2
  k = 1
  while i <= n:
     if n%i == 0:
        k *= i
     while n%i == 0:
        n /= i
     i += 1
     if k == 1 and i > math.sqrt(n):
        k *= n
        break
  return k

f = open("abcdata.txt","w")
for n in range(2,101,2):
  for m in range(1,101,2):
     a = 2*m*n
     c = m*m+n*n
     v = gcd(a,c)
     if v > 1:
        continue
     A = (c-a)*(c-a)
     B = 4*a*c
     C = (c+a)*(c+a)
     quality = math.log(C)/math.log(rad(c-a)*rad(a)*rad(c)*rad(c+a))
     if quality > 1:
        f.write(str(A)+" "+str(B)+" "+str(C)+" "+str(quality)+"\n")

Edited June 21, 2013 by phillip1882

[Turtle]

just for the "fun" of it, i wrote a program to produce a couple hundred Pythagorean triples that lead to abc conjecture "liars" yes turtle, my term for cases where its "not true".

abcdata.txt

here's the code itself.

 

import math

def gcd(a,b ):
  if a == 0:
     return b
  if b == 0:
     return a
  if a < b:
     return gcd(a,b%a)
  else:
     return gcd(a%b,b )

def rad(n):
  i = 2
  k = 1
  while i <= n:
     if n%i == 0:
        k *= i
     while n%i == 0:
        n /= i
     i += 1
     if k == 1 and i > math.sqrt(n):
        k *= n
        break
  return k

f = open("abcdata.txt","w")
for n in range(2,101,2):
  for m in range(1,101,2):
     a = 2*m*n
     c = m*m+n*n
     v = gcd(a,c)
     if v > 1:
        continue
     A = (c-a)*(c-a)
     B = 4*a*c
     C = (c+a)*(c+a)
     quality = math.log(C)/math.log(rad(c-a)*rad(a)*rad(c)*rad(c+a))
     if quality > 1:
        f.write(str(A)+" "+str(B)+" "+str(C)+" "+str(quality)+"\n")

 

 

 

 

Got it. Here's the skinny on the distributed effort.

 

 

ABC at Home Data

 

Data collected so far

We found 23652770 ABC triples so far.

 

In March 2011 we almost completed our exhaustive search for all ABC triples (a,b,c) with a < b < c < 10^18. In November 2011 we discovered that some parts of the search space were not covered and we found 6 more triples. The total number is 14482065, and we are still running final checks to make sure this list is complete. The compressed file triples_below_1018_revised.gz (178.76 MB, 2011-11-07 20:55) contains the numbers c and a on a separate line for each ABC triple a+b=c we found. Please note that, for now, this data is provided on a preliminary basis since we are not fully sure it is complete. See this table for counts according to size and quality.

 

We are currently continuing with a slightly bigger search for all ABC triples (a,b,c) with c no more than 2^63. The 9170705 triples with c at least 10^18 that we found so far are in the file big_triples.gz (130.05 MB, 2013-06-20 23:36 UTC)

Daily totals

 

 

This is a list of the number of triples found each day. The wide fluctuations are due to the fact that certain areas in the search space are more rich in triples than others.

 

2013-06-20 5333

 

2013-06-18 8229

 

...

 

I'm dowloading their files & will try and decompress them & have a look this evening.

Edited June 21, 2013 by Turtle

[Turtle]

Mind you this is new to me & I'm juggling as fast as I can. I was able to extract the data files from ABC Home and open them in Wordpad. I'll just post in quotes the first few elements of Phillip's .txt file and the first few elements of the 2 files from Home.I do not yet grok any of them.

 

Note the ABC Home files I extracted have no extension and are opening in Wordpad in a sketchy manner. I couldn't view them using the usual suspects so maybe y'all know something I dunt. The files are single spaced but they are pasting here double & I'm having no luck editing them. :shrug: You know what to do. :hammer:

 

1 80 81 1.29203002988

1 624 625 1.07904681719

6561 22000 28561 1.22665541555

279841 251600 531441 1.01560183141

390625 316656 707281 1.00774705999

531441 392080 923521 1.0758585762

81 544 625 1.03261590116

1 2400 2401 1.45567310018

1 6560 6561 1.23530331842

2401 4160 6561 1.1108234475

1 50624 50625 1.07521026401

1 83520 83521 1.18029574331

2401 277440 279841 1.10835823696

6561 384064 390625 1.06401390599

14641 516800 531441 1.13877842947

28398241 34344000 62742241 1.11169628598

1 130320 130321 1.02023733758

2401 81120 83521 1.05468295643

1 390624 390625 1.09930999132

1 531440 531441 1.08037941587

2401 388224 390625 1.15297606362

625 530816 531441 1.12377186998

81 707200 707281 1.10705714175

1 923520 923521 1.05571393994

1 1185920 1185921 1.04971431794

6561 2819200 2825761 1.06897474957

 

9 1

 

32 5

 

49 1

 

64 1

 

81 1

 

81 32

 

125 4

 

128 3

 

225 1

 

243 1

 

245 2

 

250 7

 

256 13

 

256 81

 

289 1

 

343 100

 

375 32

 

512 5

 

512 169

 

513 1

 

539 27

 

625 1

 

625 49

 

625 81

 

676 1

 

729 1

 

729 25

 

 

1000000000000000000 1

 

1000000000000000000 16303

 

1000000000000000000 94249

 

1000000000000000000 132651

 

1000000000000000000 758889

 

1000000000000000000 3944539

 

1000000000000000000 6704339

 

1000000000000000000 13329199

 

1000000000000000000 17658001

 

1000000000000000000 23778181

 

1000000000000000000 48825897

 

1000000000000000000 93103201

 

1000000000000000000 217238121

 

1000000000000000000 1999999999

 

1000000000000000000 4912918759

 

1000000000000000000 6761990971

 

1000000000000000000 7628479857

 

1000000000000000000 7988005999

 

1000000000000000000 13841287201

 

1000000000000000000 14763478339

 

 

 

 

 

EDIT: PS You might find this old thread of interest P. An algorithm to generate all the perfect right triangles

Edited June 22, 2013 by Turtle

[phillip1882]

i was well aware of the algorithm presented there, though thanks for the link none the less.

i have a hard time believing that there are finitely many cases where the abc conjecture is false for another reason as well.

take the set of primes between 1 and n.

split them into two groups. raise to powers, multiply, subtract group A from group B.

if even a few primes are missing from the group, the result is almost certainly largely composite.

[Turtle]

i was well aware of the algorithm presented there, though thanks for the link none the less.

i have a hard time believing that there are finitely many cases where the abc conjecture is false for another reason as well.

take the set of primes between 1 and n.

split them into two groups. raise to powers, multiply, subtract group A from group B.

if even a few primes are missing from the group, the result is almost certainly largely composite.

 

You're welcome, of course. :)

 

I guess if you have a proof that pertains, you can try submitting it. As I read the conjecture in its fundamental form, it does not make the finite qualification but only says [depending on who you read it from] that there are "not many" or "few" d's > c. Inspite of all the fuss made about the abc conjecture & its implications if prove/disproven, I think it's kinda a weaselly weanie conjecture. Show some brass you conjecturers!! Give me a less than 36% or some such a matter that's meaty enough to poke a forked stick at. :lol:

Edited June 29, 2013 by Turtle

[CraigD]

Educational thread, Phillip :thumbs_up It’s been my introduction to the ABC conjecture – thanks.

 

i was well aware of the algorithm presented there, though thanks for the link none the less.

i have a hard time believing that there are finitely many cases where the abc conjecture is false for another reason as well.

take the set of primes between 1 and n.

split them into two groups. raise to powers, multiply, subtract group A from group B.

if even a few primes are missing from the group, the result is almost certainly largely composite.

 

Inspite of all the fuss made about the abc conjecture & its implications if prove/disproven, I think it's kinda a weaselly weanie conjecture.

I disagree. I find it bizarre and counterintuitive.

 

Consider this obvious consequence:

 

If there are a finite number of triples [imath](a, b, c)[/imath] such that [math]a + b = c[/math] and [math]c > \left( \prod_{p\mid a}p \prod_{p\mid b}p \prod_{p\mid c}p \right)^2[/math] where [imath]p[/imath] is prime

then there must be a greatest [imath]c[/imath], [imath]c_{\mbox{max}}[/imath].

 

I believe that ABC conjecture liars with [imath]c \dot > 10^{20}[/imath] have been found through brute force computation, defining a lower bound for [imath]c_{\mbox{max}}[/imath].

 

I find the idea that, above some integer [imath]c_{\mbox{max}}[/imath], arithmetic begins behaving qualitatively differently, deeply weird.

[Turtle]

...snip

 

As I read the conjecture in its fundamental form, it does not make the finite qualification but only says [depending on who you read it from] that there are "not many" or "few" d's > c. Inspite of all the fuss made about the abc conjecture & its implications if prove/disproven, I think it's kinda a weaselly weanie conjecture. Show some brass you conjecturers!! Give me a less than 36% or some such a matter that's meaty enough to poke a forked stick at. :lol:

 

 

 

I disagree. I find it bizarre and counterintuitive.

 

Consider this obvious consequence:

 

If there are a finite number of triples [imath](a, b, c)[/imath] such that [math]a + b = c[/math] and [math]c > \left( \prod_{p\mid a}p \prod_{p\mid b}p \prod_{p\mid c}p \right)^2[/math] where [imath]p[/imath] is prime

then there must be a greatest [imath]c[/imath], [imath]c_{\mbox{max}}[/imath].

 

I believe that ABC conjecture liars with [imath]c \dot > 10^{20}[/imath] have been found through brute force computation, defining a lower bound for [imath]c_{\mbox{max}}[/imath].

 

I find the idea that, above some integer [imath]c_{\mbox{max}}[/imath], arithmetic begins behaving qualitatively differently, deeply weird.

 

 

The conjecture does not propose a finite set; that was an invention of Phillip's. Quoting from Wiki, emphasis mine:

...The abc conjecture (also known as Oesterlé–Masser conjecture) is a conjecture in number theory, first proposed by Joseph Oesterlé (1988) and David Masser (1985) as an integer analogue of the Mason–Stothers theorem for polynomials. The conjecture is stated in terms of three positive integers, a, b and c (hence the name), which have no common factor and satisfy a + b = c. If d denotes the product of the distinct prime factors of abc, the conjecture essentially states that d is usually not much smaller than c. In other words: if a and b are composed from large powers of primes, then c is usually not divisible by large powers of primes. The precise statement is given below. ...

 

It's this "usually" that I find weaselly. Bad arithmetic! :naughty: Hold still! Whips & chains aside, our arithmetic actually doesn't jump around, i.e. it is internally consistent & kurt's velvet hammer informs us such conjectures as abc may be one of those truths for which no proof is possible. :hammer:

[CraigD]

The conjecture does not propose a finite set; that was an invention of Phillip's. Quoting from Wiki ...

I’m new to the ABC conjecture (or if not, too disorderly to remember I’ve considered it before), but we appear to not be reading from the same wikipedia page (well, I'm near certain we are, so I mean not reading the same part of it) on it. Wikipedia says

ABC Conjecture. For every ε > 0, there exist only finitely many triples (a, b, c) of positive coprime integers, with a + b = c, such that

[math]c>\\rad(abc)^{1+\varepsilon}[/math]

Where the rad() function is the same as the products I used in my writing of it.

 

Wolfram Mathworld’s article doesn’t mention the “finitely many triples” version of the conjecture either, so maybe its an error injected not only into this thread, but into the wikipedia article, too. For my appreciation of it, I’d like to get the rightness or wrongness of this phrasing settled ASAP, as I haven’t gotten as far as even thinking much about the other phrasings. I’ve skimmed a couple of references from the wikipedia page, however, and that phrasing appears accurately state an accepted corollary of the conjecture.

 

I’m not sure what to call this theorem, but if a ordered set has a finite number of members, it has a last, or maximum one, so if the “finitely many triples” phrasing of the conjecture is correct, so’s my “c_max” one.

[Turtle]

I’m new to the ABC conjecture

 

Ditto.

 

(or if not, too disorderly to remember I’ve considered it before),

 

Ditto.

 

but we appear to not be reading from the same wikipedia page (well, I'm near certain we are, so I mean not reading the same part of it) on it.

 

I thought I put in the link, but see now I di'nt. Here it is: abc conjecture @ Wikipedia

 

Wikipedia says

ABC Conjecture. For every ε > 0, there exist only finitely many triples (a, b, c) of positive coprime integers, with a + b = c, such that

[math]c>\\rad(abc)^{1+\varepsilon}[/math]

Where the rad() function is the same as the products I used in my writing of it.

 

Wolfram Mathworld’s article doesn’t mention the “finitely many triples” version of the conjecture either, so maybe its an error injected not only into this thread, but into the wikipedia article, too. For my appreciation of it, I’d like to get the rightness or wrongness of this phrasing settled ASAP, as I haven’t gotten as far as even thinking much about the other phrasings. I’ve skimmed a couple of references from the wikipedia page, however, and that phrasing appears accurately state an accepted corollary of the conjecture.

 

I’m not sure what to call this theorem, but if a ordered set has a finite number of members, it has a last, or maximum one, so if the “finitely many triples” phrasing of the conjecture is correct, so’s my “c_max” one.

 

I just don't know enough about it to make much of anymore noise. I'll try & do some more reading, starting with this piece I just rooted out. >> The A B C's of Number Theory by Prof. Noam D. Elkies Harvard University

 

A taster...

 

Abstract

 

The ABC conjecture is a central open problem in modern number theory, connecting results, techniques and questions ranging from elementary number theory and algebra to the arithmetic of elliptic curves to algebraic geometry and even to entire functions of a complex variable. The conjecture asserts that, in a precise sense that we specify later, if A,B,C are relatively prime integers such that A + B = C then A,B,C cannot all have many repeated prime factors.

 

Elliptic curves and complex variables...yay! :huh: (Don't know what's up with that abstract quote formatting, but I give!! :wacko: )

Edited June 30, 2013 by Turtle

[Turtle]

...I just don't know enough about it to make much of anymore noise. I'll try & do some more reading, starting with this piece I just rooted out. >> The A B C's of Number Theory by Prof. Noam D. Elkies Harvard University

 

 

 

I read Elkie's paper and found much of it beyond my grokkeage if not kenning. Nonetheless, he makes no indication that there is anything of the abc conjecture which would cause Arithmetic to take a knee.

[phillip1882]

the abc conjecture is generally "false" around powers of primes. that is, a b and c is a power of a prime or two.

in order for the abc conjecture to be true, there's a point at which powers of primes behave differently than they did before. essentially you would be unable to get another power of a prime from subtraction/addition. this is what Craig is referring to.

here would be my abc lair candidate set:

1, 4, 8, 9, 16, 24, 25, 27, 32, 36, 48, 49, 54, 64, 80, 81, 96, 100, 108, 112, 121,

125, 128, 135, 144, 160, 162, 169, 189, 192, 196, 216, 224, 225, 243, 250, 256,

288, 289, 320, 324, 343, 352, 361, 375, 384, 400, 405, 416, 432, 441, 448, 480,

484, 486, 500, 512, 529, 567, 576, 625, 640, 648, 676, 686, 704, 729, 750, 768,

784, 810, 832, 841, 864, 875, 891, 896, 900, 960, 961, 972, 1000, 1024, 1029,

1053, 1080, 1088, 1089, 1125, 1134, 1152, 1156, 1215, 1216, 1225, 1250, 1280,

1296, 1331, 1344, 1369...

despite the fact this set goes into infinity, there are only finitely many cases where you can produce a third number from the sum of two others; where all three numbers are co-prime.

 

and here would be the code that produced it.

powers = []
i = 2
while i*i < 10000:
  powers += [i*i]
  k = powers[-1]
  while k*i<10000:
     powers += [k*i]
     k = powers[-1]
     j = 2
     v = powers[-1]/i
     while j < v:
        powers += [j*k]
        j += 1
  i+=1
powers.sort()
i = 0
while i+1 < len(powers):
  if powers[i] == powers[i+1]:
     powers.pop(i)
  else:
     i += 1
print powers[:100]