= abc conjecture
{c}
{title2=$C\leq K_\epsilon\operatorname{rad}(ABC)^{1+\epsilon}$}
For every $\epsilon>0$ there is a constant $K_\epsilon$ 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 $ax^n-by^n=c\ne0$, division by the gcd, which divides $c$, gives a uniform bound on $\max(x,y)^{n-2(1+\epsilon)}$. Finiteness in the varying exponent requires excluding $x=y=1$ when $a-b=c$.
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