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Problem: Let $p,q,r$, be positive integers satisfying $\frac {1}{p} + \frac {1}{q} + \frac {1}{r} < 1$ . If the ABC conjecture is true, then $x^p + y^q = z^r$ has finitely many positive integer solutions $(x,y,z)$ that are co-prime.

Thoughts:

1) $p,q,r> 1$ because of the inequality.

2) In the formulation of the ABC conjecture I am familiar with , it requires that $gcd(a,b,c)=1$, so in our case I might need to show that $gcd(x^p, y^q,z^r)=1$

3) I know that if $gcd(x,y,z)=1$ then $rad(x^p, y^q,z^r)=rad(x,y,z)$

other than that I am not sure how to proceed. Insights appreciated.

IntegrateThis
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  • As it stands now, your statement is trivially wrong. It must have contained some words along the lines of "coprime" or "primitive". Short of that, let $(p,q,r)=(7,3,2)$ and have the solutions $(x,y,z)=(1,2,3)$, then $(2^6\cdot1,2^{14}\cdot2,2^{21}\cdot3)$, then $(3^6\cdot1,3^{14}\cdot2,3^{21}\cdot3)$, then $(4^6\cdot1,4^{14}\cdot2,4^{21}\cdot3)$, and so on. – Ivan Neretin Feb 08 '19 at 09:02
  • @IvanNeretin I think there was a mistake, for now I have changed the problem so that $x,y,z$ are co-prime. – IntegrateThis Feb 08 '19 at 17:18
  • Then you have your $\gcd(x,y,z)=1$ and the abc conjecture fits like a glove, so what's the question now? – Ivan Neretin Feb 08 '19 at 18:02
  • @IvanNeretin still new to this topic, will post an answer later if I understand thanks. – IntegrateThis Feb 08 '19 at 19:40

1 Answers1

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Beukers in his 2005 lecture gives the following:

The ABC-conjecture

  • ABC-Conjecture (Masser-Oesterl´e, 1985): Let $\kappa > 1$. Then, with finitely many exceptions we have $C < rad(ABC)^\kappa$.

Fermat-Catalan conjecture

Consequence of $ABC$-conjecture:

The set of triples $x^p$, $y^q$, $z^r$ with $x$, $y$, $z$ coprime positive integers

such that $x^p$ + $y^q$ = $z^r$ and $1/p + 1/q + 1/r < 1$, is finite.

  • Observation, $1/p + 1/q + 1/r < 1$ implies $1/p + 1/q + 1/r \lt 1 − 1/42$.

  • Apply $ABC$ with $\kappa = 1.01$ to $A = x^p$, $B = y^q$, $C = z^r$ . Notice that $rad(x^r y^q z^r ) \le xyz < z^{r/p} z^{r/q} z$.

  • Hence, with finitely many exceptions we get $z^r < z^{\kappa(r/p+r/q+1)}$

  • This implies $r < \kappa(r/p + r/q + 1)$ and hence $1 < \kappa(1/p + 1/q + 1/r )$. But this is impossible because $\kappa = 1.01$ and $1/p + 1/q + 1/r \le 1 − 1/42$.

Pythagorus
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