3
$\begingroup$

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 :

$$2^4+3^2=5^2$$

$$2^5+7^2=3^4$$

$$2^2+11^2=5^3$$

$$2^7+17^3=71^2$$

$$7^3+13^2=2^9$$

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.

$\endgroup$
  • 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
1
$\begingroup$

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 .

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.