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$\DeclareMathOperator\rad{rad}$Conjecture: If $A, B, C$ are positive integers with $\gcd(A, B)=1$, $\gcd(B, C)=1$, and $\gcd(C, A)=1$, and if $A+B=C$, then $\min(A,B) \le \rad(ABC)$.

If the conjecture is valid, then we can use the conjecture to prove the Fermat last theorem as follows:

Proof of the Fermat last theorem:

We consider the Fermat equation:

$x^n+y^n=z^n$ with $\gcd(x, y)=1$, $\gcd(y, z)=1$, and $\gcd(z, x)=1$.

There is no loss of generality in assuming that $x \le y <z$. By the conjecture, we get $x^n \le \rad(x^n y^n z^n)=\rad(x y z)\le x y z $.

So $x^n+y^n \le xyz+y^n<z^3+(z-1)^n$.

But we can easily prove that $z^3+(z-1)^n < z^n$ whenever $n > 3$ and $z>1$. So now we only need to prove the Fermat last theorem with $n=3$.

My question: Is the conjecture above new and correct?

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    $\begingroup$ Setting aside the fact that a counterexample was produced in an hour, surely MO is not the place to test conjectures that trivially imply FLT. It seems to me that, for such a conjecture, you need to provide some serious evidence that it might be correct before expecting other people to spend time on it. $\endgroup$
    – LSpice
    Commented Aug 23, 2019 at 18:50
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    $\begingroup$ @LSpice, to be fair, it is true when $a, b \leq 1000.$ $\endgroup$
    – J. E. Pascoe
    Commented Aug 23, 2019 at 18:54
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    $\begingroup$ In line with the counterexample given by Pascoe, looking for differences and sums of large powers of small primes that only have a few distinct small prime factors probably gives many counterexamples. $\endgroup$
    – Robin Zhang
    Commented Aug 23, 2019 at 19:03
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    $\begingroup$ Not that I want to encourage naive conjecture here, but it seems to me that the conjecture seems to correspond to taking $\varepsilon = 0$ in the $abc$-conjecture, but correspondingly weakening the conclusion to be about the smallest number involved rather than the largest. So far, the Mathematica program I have been running has produced one other example with $a,b< 10000,$ which is $(1024,1377).$ Its possible there may be only finitely many counterexamples, but much like the $abc$-conjecture itself, I see no reason why. $\endgroup$
    – J. E. Pascoe
    Commented Aug 23, 2019 at 19:10
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    $\begingroup$ I'm voting to close this question for the reasons indicated by @LSpice $\endgroup$
    – Yemon Choi
    Commented Aug 24, 2019 at 3:26

1 Answer 1

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Note, $$625+2048=5^4 + 2^{11} = 3^5\times 11 = 2673.$$ The relevant radical is thus $$2\times 3 \times 5 \times 11=330.$$ Therefore, the conjecture is false.

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    $\begingroup$ The complete list of $(a,b)$ where $a \leq b < 10000$ which violate the conjecture are $(625,2048), (1024,1377), (8019,8788), (8192,8575).$ $\endgroup$
    – J. E. Pascoe
    Commented Aug 23, 2019 at 19:40
  • $\begingroup$ Can You help me check the conjecture 1 at mathoverflow.net/questions/338117 with $0< a \le b \le 10^6$ $\endgroup$
    – Đào Thanh Oai
    Commented Aug 24, 2019 at 5:44
  • $\begingroup$ I can tell you that a back of the envelope calculation (probably) shows that if the $abc$-conjecture is true, then there are finitely many counterexamples. However, naive conjectures are not appropriate for mathoverflow. Problems that obviously imply big theorems should be worked on in private. The only reason I spent the 5 minutes to code this is that I thought there was some novelty in replacing $c$ by $a$ is the statement of the $abc$-conjecture. Any proof of any of the conjectures you have posted would require deep mathematics, and would not be suitable for a quick mathoverflow answer. $\endgroup$
    – J. E. Pascoe
    Commented Aug 24, 2019 at 12:37
  • $\begingroup$ I have some ideas, I want private communication with you via email. But I don’t know your email, what is your email? $\endgroup$
    – Đào Thanh Oai
    Commented Aug 24, 2019 at 13:11

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