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a simple disproof of the abc conjecture

Posted: Sun Sep 16, 2018 6:18 pm UTC
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

Re: a simple disproof of the abc conjecture

Posted: Sun Sep 16, 2018 8:15 pm UTC
Please state this in a way that's actually comprehensible.

Re: a simple disproof of the abc conjecture

Posted: Sun Sep 16, 2018 8:35 pm UTC
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*32, 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 Kε such that if a,b,c are natural numbers with a+b = c, c < Kε*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:

9n-1 is always a multiple of 8 for any positive integer n, and rad(9n) = 3, since 9n = 32n has only 3 as a prime factor. For some n, consider
kn = (9n-1)/8. Also consider the ratio 9n/(6k).

Re: a simple disproof of the abc conjecture

Posted: Sun Sep 16, 2018 8:46 pm UTC
I mean, I guess a = 1, b = 9n-1, and c = 9n. Then rad(abc) = rad(9n(9n-1)) = 3 rad(9n-1) ≥ 6. The last inequality holds because 9n-1 is divisible by 8 = 23. Of course, if n > 1, it must have other prime factors as well, since 9n-1 is never a power of 2 unless n = 1, and since none of the other factors is 3 (9n-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.

Re: a simple disproof of the abc conjecture

Posted: Sun Sep 16, 2018 8:53 pm UTC
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
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.

Re: a simple disproof of the abc conjecture

Posted: Sun Sep 16, 2018 10:03 pm UTC
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.

Re: a simple disproof of the abc conjecture

Posted: Sun Sep 16, 2018 10:04 pm UTC
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.

Re: a simple disproof of the abc conjecture

Posted: Sun Sep 16, 2018 10:05 pm UTC
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.

Re: a simple disproof of the abc conjecture

Posted: Mon Sep 17, 2018 12:10 am UTC
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.

Re: a simple disproof of the abc conjecture

Posted: Mon Sep 17, 2018 10:41 am UTC
The quality is log(c)/log(rad(abc)) = log(9n)/log(rad(92n-9n)). The denominator might feel like it ought to approach log(rad(92n-9n+¼)) = log(rad((9n-½)2)) = log(rad(9n-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.

Re: a simple disproof of the abc conjecture

Posted: Mon Sep 17, 2018 7:00 pm UTC
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.