abc conjecture ($C\le K_{ϵ}\mathrm{rad}(ABC{)}^{1+ϵ}$)

For every $ϵ>0$ there is a constant $K_{ϵ}$ such that the displayed inequality holds for coprime positive integers $A+B=C$. The radical is the product of the distinct primes dividing the argument. Applying it to a primitive Fermat triple bounds the exponent and the triple simultaneously. For $a{x}^n-b{y}^n=c\ne 0$, division by the gcd, which divides $c$, gives a uniform bound on $\max (x,y{)}^{n-2(1+ϵ)}$. Finiteness in the varying exponent requires excluding $x=y=1$ when $a-b=c$.

Put $M=x^p$ for positive bases $x,y\ge 2$. The coprime triple $(y^q,1,x^p)$ has radical at most $xy<M^{2/3}$. The abc conjecture with $ϵ=1/6$ gives $M\le C_{1/6}{M}^{7/9}$, proving the bound. Both bases and both exponents are then bounded, since $2^p\le M$ and $2^q\le M-1$. Unlike a homogeneous Fermat equation, no common-scaling ambiguity occurs.

For a primitive positive solution of $x^n+y^n=z^n$ with $n\ge 4$, the three powers are pairwise coprime and their radical is at most $xyz<z^3$. The abc conjecture with $ϵ=1/6$ gives the displayed inequality, hence $z\le C_{1/6}^2$ uniformly in $n$. The inequality $z^n\le 2\max (x,y{)}^n$ then bounds $n$. Thus there are finitely many primitive triples and exponents. Homogeneity would produce infinitely many unscaled solutions from any one nonzero solution, so a primitive arithmetic height bound alone must not be described as finite many arbitrary multiples.