Let $p,q,r$ be distinct primes and $a,b,c\ge 2$ integers.

The equation $$p^a+q^b=r^c$$ has the following solutions :






Can we prove without using any unproven conjecture that this list is complete ? If not, does the explicit abc-conjecture imply that the list is complete ?

With the explicit abc-conjecture, I mean the following :

Let $a,b$ be coprime positive integers not both $1$, let $c=a+b$ , let $n=rad(abc)$ be the product of the distinct prime factors of $abc$ and $\omega=\omega(n)$ the number of distinct prime factors of $n$. Then, the inequality $$c<\frac{6}{5}\cdot n\cdot \frac{(\ln(n))^{\omega}}{\omega!}$$ holds.

We can assume $p<q$ and one of the primes must be $2$ because otherwise the left side would be even and the right side odd.

  • 1
    $\begingroup$ mathworld.wolfram.com/BealsConjecture.html $\endgroup$ – Jose Arnaldo Bebita-Dris Apr 22 '18 at 11:17
  • $\begingroup$ @JoseArnaldoBebitaDris This conjecture covers only the case that all exponents are at least $3$, so even assuming this we have not shown that the list is complete. If we assume that the catalan-conjecture is true in the strong sense (we know all solutions), then only a few cases are remaining. $\endgroup$ – Peter Apr 22 '18 at 11:37
  • $\begingroup$ Moreover, I am not sure whether the EXPLICIT abc-conjecture implies Beal's and/or Catalan/Fermat's conjecture. $\endgroup$ – Peter Apr 22 '18 at 11:39
  • 2
    $\begingroup$ It's hard to prove even single case, $2^n=p^3+q^2$ for me now... $\endgroup$ – didgogns Apr 23 '18 at 16:48
  • $\begingroup$ This post may be of interest. $\endgroup$ – Tito Piezas III Apr 23 '18 at 17:00

In this answer, I'll assume Explicit abc-Conjecture is true and will show that the set of solution is finite (the upper bound is very big). Since exactly one of $p$, $q$ and $r$ is $2$, we can split the equation into two cases.

i) $r=2$.

The equation is now $p^a+q^b=2^c$. When $a=b=2$, $p^a+q^b=p^2+q^2\equiv 2\pmod 4$, therefore $c=1$ and no solution. Therefore, $\frac{1}{a}+\frac{1}{b}\le\frac{1}{2}+\frac{1}{3}=\frac{5}{6}$. Let $p^a+q^b=2^c=x$, then $p<x^{1/a}$ and $q<x^{1/b}$. It is obvious that $pq<x^{5/6}$. Therefore by Explicit abc,$$x=2^c<\frac{6}{5}\cdot2pq\cdot\frac{(\log{2pq})^3}{6}<\frac{2x^{5/6}\times\log^3(2x^{5/6})}{5}$$and solving it for $x$ gives $x<1.489\times10^{29}$.

ii) $q=2$.

The equation is now $p^a+2^b=r^c$. Similarly when $a=c=2$, the equation becomes $2^b=(r+p)(r-p)$ and one can show that $3^2+2^4=5^2$ is the only solution. Now we can assume $\frac{1}{a}+\frac{1}{c}\le\frac{5}{6}$ and if we let $p^a+2^b=r^c=x$, then $pr<x^{5/6}$. By Explicit abc,$$x=r^c<\frac{6}{5}\cdot2pr\cdot\frac{(\log{2pr})^3}{6}<\frac{2x^{5/6}\times\log^3(2x^{5/6})}{5}$$and this is exactly the same inequality as the first case.

I verified by computer that there is no other solutions assuming explicit abc conjecture. It took ~15 hours with my machine, and the code is available at https://github.com/didgogns/number_theory/blob/master/2748541.py .


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