Numbers with large prime exponents and the ABC conjecture

Question

By Fermat's Last Theorem, there are no solutions to the Diophantine equation $a^n + b^n = c^n$ for $a,b,c > 0$ and $n>2$. Beal's conjecture allows the exponents to be different (but also $>2$ ). Is the lack of solutions because there is not enough wiggle room? (Squares are too abundant, but what about cubes and so on?)

Let's call a positive integer N-power-min if the smallest exponent in its prime factorization is $N$. For $N=3$, I've downloaded a list of ABC Triples and the first solutions were:

$271^{3} + 2^{3} 3^{5} 73^{3} = 919^{3}$

and

$3^{4} 29^{3} 89^{3} + 7^{3} 11^{3} 167^{3} = 2^{7} 5^{4} 353^{3}$

Nothing with $N>3$ was found.

Does the ABC conjecture (or a related one) imply that for any $N > 2$ , we have a limited or no solutions $a+b=c$ with a,b,c N-power-min ?

Answer 1

If $a,b,c$ are $N$-power min then $\operatorname{rad}(abc) \leq (abc)^{1/N} \leq c^{ 3/N}$

and the $abc$ conjecture implies that $$c< K_\epsilon \operatorname{rad}(abc)^{1+\epsilon} \leq K_\epsilon c^{ (3/N)(1+ \epsilon)} $$ so $$ c< K_\epsilon^{ \frac{1}{ 1 - (3/N)(1+\epsilon)}}$$

But if $c$ is $N$-power-min then $c \geq 2^N$, so for $N$ sufficiently large depending on $\epsilon, K_\epsilon$, this inequality cannot be satisfied.

For $N$ sufficiently large depending on $\epsilon, K_\epsilon$, these inequalities cannot be satisfied.

So certainly the abc conjecture implies for some $N>2$, we have no solutions.

But, as stated, it doesn't imply (by any easy argument) that for every $N>2$ we have no solutions, nor does it even imply for any particular $N$ that there is no solution, since the exact value of $K_\epsilon$ is not given in the statement (as was explained by Mark Sapir in the comments).

---

There probably are infinitely many solutions for $N=3$, albeit very sparse ones. You just need to search for an elliptic curve of the form $a^4 x^3 + b^4 y^3 + c^4 z^3$ of positive rank, then find various $x,y,z$ solutions.