## a simple disproof of the abc conjecture

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- phillip1882
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### a simple disproof of the abc conjecture

9^n -1 is divisible by 8, and 9^n is 3 with the rad function. thus at worst you have 9^n/(6*k) where k is (9^n-1)/8

good luck have fun

- Xanthir
- My HERO!!!
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### Re: a simple disproof of the abc conjecture

Please state this in a way that's actually comprehensible.

(defun fibs (n &optional (a 1) (b 1)) (take n (unfold '+ a b)))

- Eebster the Great
**Posts:**3460**Joined:**Mon Nov 10, 2008 12:58 am UTC**Location:**Cleveland, Ohio

### Re: a simple disproof of the abc conjecture

I don't know, but I'll give some background in case you or someone else looking at the thread wants it.

He is referencing the "abc conjecture," an open problem in number theory. The conjecture deals with the radical function rad(n), which returns the product of distinct prime factors of n. For instance, since 18 = 2*3

phillip is claiming to have proved this longstanding conjecture false in a single line, but I admit I don't get it either. Here's as far as I got:

9

k

That's about it.

He is referencing the "abc conjecture," an open problem in number theory. The conjecture deals with the radical function rad(n), which returns the product of distinct prime factors of n. For instance, since 18 = 2*3

^{2}, it has two distinct prime factors, 2 and 3, so rad(18) = 2*3 = 6. The conjecture specifically deals with three natural numbers, a, b, and c, with a+b=c. It turns out that in this case, rad(abc) is usually less than c itself, but not always. However, if we take rad(abc) to some power greater than 1, there will only be finitely many cases (if any) where c > rad(abc). One form of the conjecture states:For any real ε > 0, there exists a natural number K_{ε}such that if a,b,c are natural numbers with a+b = c, c < K_{ε}*rad(abc)^{1+ε}.

phillip is claiming to have proved this longstanding conjecture false in a single line, but I admit I don't get it either. Here's as far as I got:

9

^{n}-1 is always a multiple of 8 for any positive integer n, and rad(9^{n}) = 3, since 9^{n}= 3^{2n}has only 3 as a prime factor. For some n, considerk

_{n}= (9^{n}-1)/8. Also consider the ratio 9^{n}/(6k).That's about it.

- Eebster the Great
**Posts:**3460**Joined:**Mon Nov 10, 2008 12:58 am UTC**Location:**Cleveland, Ohio

### Re: a simple disproof of the abc conjecture

I mean, I guess a = 1, b = 9

^{n}-1, and c = 9^{n}. Then rad(abc) = rad(9^{n}(9^{n}-1)) = 3 rad(9^{n}-1) ≥ 6. The last inequality holds because 9^{n}-1 is divisible by 8 = 2^{3}. Of course, if n > 1, it must have other prime factors as well, since 9^{n}-1 is never a power of 2 unless n = 1, and since none of the other factors is 3 (9^{n}-1 is clearly not divisible by 3, since it is one less than a power of 3), we have a strict inequality whenever n > 1. Not sure how this is useful, though.- phillip1882
**Posts:**145**Joined:**Fri Jun 14, 2013 9:11 pm UTC**Location:**geogia-
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### Re: a simple disproof of the abc conjecture

okay i'll go over the whole thing.

it is conjectured that given A+B = C where the three values share no common factor when reduced to just the unique primes,

multiplied together,will be less than C a finitely many times if raising that product to a power >1.

for example 32 +49 = 81

the unique factors are 2 7 and 3, and 81 > (2*7*3).

log (81) / log(42) is the quality of the result. this is roughly equal to 1.176. which means 42 can be raised to the power of 1.176 for it to be greater than 81. the 2 7 and 3 are the radical. the conjecture states that given a fixed value for the power greater than 1 will result in finitely many A+B= C

where rad(A*B*C)^k < C .

however with 1 +(9^n-1) = 9^n, 9^n-1 is divisible by 8, and 9^n is just 3, which means for any integer value of n 1 or more will result in 9^n /rad(2*3*(9^n-1)/8) which means the power can be as high as 1.026 approximately before it's larger.

it is conjectured that given A+B = C where the three values share no common factor when reduced to just the unique primes,

multiplied together,will be less than C a finitely many times if raising that product to a power >1.

for example 32 +49 = 81

the unique factors are 2 7 and 3, and 81 > (2*7*3).

log (81) / log(42) is the quality of the result. this is roughly equal to 1.176. which means 42 can be raised to the power of 1.176 for it to be greater than 81. the 2 7 and 3 are the radical. the conjecture states that given a fixed value for the power greater than 1 will result in finitely many A+B= C

where rad(A*B*C)^k < C .

however with 1 +(9^n-1) = 9^n, 9^n-1 is divisible by 8, and 9^n is just 3, which means for any integer value of n 1 or more will result in 9^n /rad(2*3*(9^n-1)/8) which means the power can be as high as 1.026 approximately before it's larger.

good luck have fun

- phillip1882
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### Re: a simple disproof of the abc conjecture

okay i thought that this was true, but after testing some values, i find that power does approach 1 the larger you go. i'm not certain this is a valid disproof now.

good luck have fun

- Xanthir
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### Re: a simple disproof of the abc conjecture

however with 1 +(9^n-1) = 9^n, 9^n-1 is divisible by 8, and 9^n is just 3, which means for any integer value of n 1 or more will result in 9^n /rad(2*3*(9^n-1)/8) which means the power can be as high as 1.026 approximately before it's larger.

You need to go into this more. What precisely do you think you've proven here?

As Eebster said, in triples of the form you outline, the radical is `rad(9ⁿ * 9ⁿ-1 * 1)`, which simplifies to `3 * rad(9ⁿ-1)`; at n=1 that equals 6, at n=2 that equals 30 (rad(80) = 2 * 5), at n=3 that equals 546 (rad(728) = 2 * 7 * 13), at n=4 that equals 1230 (rad(6560) = 2*5*41), etc.

At these values rad(abc) is indeed less than c. You'll have to further explain the leap that you're making where rad(abc)ⁱ is less than c infinitely many times, for any fixed ⁱ > 1.

(defun fibs (n &optional (a 1) (b 1)) (take n (unfold '+ a b)))

- Xanthir
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### Re: a simple disproof of the abc conjecture

Ah, ninja'd.

Yeah, looking at a few values does not disprove anything. What you have so far is not remotely a disproof, it's an observation that this particular pattern generates "big Cs" more often than other patterns of similar complexity.

Yeah, looking at a few values does not disprove anything. What you have so far is not remotely a disproof, it's an observation that this particular pattern generates "big Cs" more often than other patterns of similar complexity.

(defun fibs (n &optional (a 1) (b 1)) (take n (unfold '+ a b)))

- phillip1882
**Posts:**145**Joined:**Fri Jun 14, 2013 9:11 pm UTC**Location:**geogia-
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### Re: a simple disproof of the abc conjecture

its fairly straight forward, in general a^n -1 is divisible by a-1.

so 9^n -1 is divisable by 8.

thus you basically have(9^n)/(3/4 *(9^n -1)) . where i'm lost is that i thought this would always give a quality of approximately log(4)/log(3) but that doesn't seem to be the case.

so 9^n -1 is divisable by 8.

thus you basically have(9^n)/(3/4 *(9^n -1)) . where i'm lost is that i thought this would always give a quality of approximately log(4)/log(3) but that doesn't seem to be the case.

good luck have fun

- Eebster the Great
**Posts:**3460**Joined:**Mon Nov 10, 2008 12:58 am UTC**Location:**Cleveland, Ohio

### Re: a simple disproof of the abc conjecture

The quality is log(c)/log(rad(abc)) = log(9

^{n})/log(rad(9^{2n}-9^{n})). The denominator might feel like it ought to approach log(rad(9^{2n}-9^{n}+¼)) = log(rad((9^{n}-½)^{2})) = log(rad(9^{n}-½)), but that's not really how the rad function works, and also, I think this isn't quite going where you wanted anyway. You need to think through these problems a little longer, and expect that decades of brilliant mathematicians probably haven't missed anything obvious.- phillip1882
**Posts:**145**Joined:**Fri Jun 14, 2013 9:11 pm UTC**Location:**geogia-
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### Re: a simple disproof of the abc conjecture

i tried a slight more complex result:

9^(2^n) -1 is divisible by 2^n, but even this reduces to near 1 over larger n values for the quality.

9^(2^n) -1 is divisible by 2^n, but even this reduces to near 1 over larger n values for the quality.

good luck have fun

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