It is well known that the ABC conjecture gives an immediate proof of Fermat Last Theorem (FLT). It seems that it proves something stronger involving not necessarily perfect power, which may still be open.

We first observe there is a natural extension of the notion of a perfect power of an integer to all positive integers. We define the average power of a positive integer $m$ to be $ap(m):=\frac{\log(m)}{\log(rad(m))}$, where $rad(m)$ is the product of primes dividing $m$. For positive integers $m,k$, we have $ap(m^k)=k.ap(m) \ge k$ with equality iff $m$ is squarefree. $ap(3^2.5.7)>1$ but is less than $ap(3.5.7^2)$. In fact every positive integer $m$ has a unique representation as $m=m_1m_2^2m_3^3...$ where $m_j$ is the product of primes $p$ such that $p^j$ exactly divides $m$ so that $ap(m)=\frac{\sum j \log(m_j)}{\sum \log(m_j)}$ is indeed the correctly weighted average of individual powers in $m$.

We shall say that a triple $(a,b,c)$ of positive integers is a abc tuple if $c=a+b$ and $gcd(a,b)=1$. The ABC conjecture states that for every $\epsilon >0$, and any abc tuples with at most finitely many exceptions, $c <rad(abc)^{1+\epsilon}$. This implies $abc \le 4abc \le c^3 < rad(abc)^{3+3\epsilon}$. So that we have $ap(abc)<3+\epsilon$ for every abc tuple with at most finitely many exception. It follows that the condition $$minap:=min\{ap(a),ap(b),ap(c)\} \ge 3+\epsilon,$$ which clearly implies $ap(abc) \ge 3+ \epsilon$, can only holds for at most finitely many abc tuples.

An effective version(no exception) of ABC for $\epsilon=1$ would imply that there is no abc tuple with $a,b,c$ each of average power $\ge 4$, and this is an enhanced version of FLT since it applies to all abc tuples, not just tuple of perfect powers. One natural question is does this enhanced version of FLT follow from the extension of Taylor and Wiles' approach? It can be phrased as an additive problem : that the set $\{m: ap(m) \ge 4\}$ is sum free and apparently cannot be phrased in terms of finitely many Diophantine equations.

The known abc tuple $$2^5.11^2.19^9+5^{15}.37^2.47=3^7.7^{11}.743, ap(abc)=4.2680083..., minap=3.69810313...$$ suggests that $4$ may be the correct power.