Is this set of solutions complete?

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.

• mathworld.wolfram.com/BealsConjecture.html – Jose Arnaldo Bebita-Dris Apr 22 '18 at 11:17
• @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. – Peter Apr 22 '18 at 11:37
• Moreover, I am not sure whether the EXPLICIT abc-conjecture implies Beal's and/or Catalan/Fermat's conjecture. – Peter Apr 22 '18 at 11:39
• It's hard to prove even single case, $2^n=p^3+q^2$ for me now... – didgogns Apr 23 '18 at 16:48
• This post may be of interest. – 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 $$p and $$q. It is obvious that $$pq. Therefore by Explicit abc,$$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. 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.