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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... 2