Download If I pick some random

Here’s a different formulation of the abc conjecture, slightly modified to make it
easier to understand: If I pick some random number e > 0 then there will always
e+1
exist another number D such that if a + b = c then max (a, b, c) ≤ D (abc)
Before explaining this I’ll rewrite it with actual numbers: If I pick some random
number, say 2, which is greater than zero then there will always exist another
number D, say 1 in this case, such that if 3 + 4 = 7 then
max (3, 4, 7) ≤ C (3 ∗ 4 ∗ 7)
1+2
i.e.
max (3, 4, 7) ≤ 1 ∗ (84)3 ,
i.e.
7 ≤ (84)3 ,
i.e.
7 ≤ 592, 704.
Here max (a, b, c) means the maximum of a, b or c, so max (3, 7, 4) = 7. To
understand (abc)1+e :
First, more or less we can say that 21 means 2 multiplied by itself once. Second
we can say that 22 = 4 can be written like 4 = 2 ∗ 2 = 22 = 21+1 . Third we look
at 42 = 16. Since 4 = 2 ∗ 2 we can rewrite 16 = 42 = (2 ∗ 2)2 .
1+e
So max (a, b, c) ≤ (abc)
just means that the biggest of your 3 numbers will
always be smaller than some multiple D of the product of the 3 numbers, abc,
as long as you raise the product abc of your 3 numbers to some power. This
multiple will depend upon the number you originally chose raise abc to.
Pick 1, 8 & 9 as our a, b & c and we get:
max (1, 8, 9) ≤ D (1 ∗ 8 ∗ 9)
1+e
⇒ 9 ≤ D ∗ (72)
1+e
If I choose e = 1 then:
(1+1)
9 ≤ D ∗ (72)
⇒ 9 ≤ D ∗ (72)2 .
So here we see that given e = 1 > 0 there exists another number D, say D = 1,
such that max (1, 8, 9) ≤ D(1 ∗ 8 ∗ 9)1+e because 9 ≤ 1 ∗ (72)2 . See here that D
is hypothesized to exist after we’ve picked our e value.
This theorem only holds if a, b & c have no common factor other than 1 (i.e. 4
& 6 have a common factor of 2 because 4 = 4 ∗ 1 = 2 ∗ 2 & 6 = 6 ∗ 1 = 2 ∗ 3 - so 4
& 6 won’t feature in this theorem together, while 5 & 6 have no common factor other
than 1 because 5 = 5 ∗ 1 & 6 = 6 ∗ 1 = 2 ∗ 3, so we could use these).
What the theorem says is that most of the prime factors of certain numbers
must occur to the first power, so in mikhail’s example of 84 = 2 ∗ 3 ∗ 7 we see
most of the prime factors have no exponent like 2 has. Further it says that
if you have small numbers, like 2 in 84 = 22 ∗ 3 ∗ 7, that are raised to higher
1
powers than 1 then there should be a big prime factor, e.g. 7, only to the first
power so as to offset the balance & compensate. So if we had 2 + 1 = k then
k would have a large prime factor as long as n is large. Picking n = 6 gives
26 + 1 = 65 = 65 ∗ 1 i.e. a large prime factor of 65... This is what the article
means more or less when talking about the ”square-free” part.
1+e
1+e
1+e
Also, note that max (a, b, c) ≤ D (abc)
implies a ≤ D (abc) , b ≤ D (abc)
1+e
& c ≤ D (abc) , so instead of a, b & c we choose an , bn &cn we can use the fact
that a + b = c is part of our hypothesis to work on an + bn = cn , i.e. Fermat’s
last theorem for relatively prime integers...
That’s all I know about it, the way the article talks about it all makes little
sense to me quite frankly, it seems like they’re discussing the above with different
algebra chosen specifically to make it awkward...
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