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For every $ϵ>0$, the abc conjecture asserts that there is a constant $K_{ϵ}>0$ such that all coprime positive integers $A+B=C$ satisfyThe radical of an integer is the product of its distinct prime factors. The condition $\gcd (A,B)=1$ makes the three entries pairwise coprime. Repeated prime powers enlarge a number without enlarging its radical in the same way, so the conjecture constrains how often addition can relate unusually concentrated prime factorizations.

The example $(A,B,C)=(1,8,9)$ has $\mathrm{rad}(ABC)=6<9$ and is an abc-hit. Such examples are compatible with the conjecture: its bound includes both the constant $K_{ϵ}$ and the exponent $1+ϵ$. Conditional applications include the primitive Fermat bound from abc and the Catalan power bound from abc.

More generally, fix positive integers $a,b$, a nonzero integer $c$, and positive integers $x,y,n$ satisfying $a{x}^n-b{y}^n=c$. Put $g=\gcd (a{x}^n,b{y}^n)$, which divides $|c|$. Applying the conjecture to the normalized positive triple formed from $a{x}^n$, $b{y}^n$ and $|c|$ gives, for $M=\max (x,y)$,Indeed, the largest normalized entry is at least a fixed multiple of $M^n$, while the radical is at most $\mathrm{rad}(ab|c|)xy\le \mathrm{rad}(ab|c|)M^2$. Taking $ϵ=1/4$ makes the exponent at least $1/2$ for $n\ge 3$, bounding $M$ uniformly; if $M\ge 2$, it also bounds $n$. The separate case $x=y=1$ can satisfy the equation for every $n$ when $a-b=c$.

The abc conjecture says that for every $ϵ>0$ there is $C_{ϵ}$ such that coprime positive integers $A+B=C$ satisfy $C\le C_{ϵ}\mathrm{rad}(ABC{)}^{1+ϵ}$, where the radical is the product of distinct prime divisors.

The Catalan alternative yields finiteness of actual nontrivial solutions, with $x,y\ge 2$ and $p,q\ge 3$. Put $M=x^p$, so $y^q=M-1$ and $x,y$ are coprime. The triple $(y^q,1,x^p)$ has radical at most $xy<M^{2/3}$. Take $ϵ=1/6$ to obtainBoth bases are bounded, and $2^p\le M$, $2^q\le M-1$ also bound both exponents. Thus only finitely many nontrivial Catalan solutions exist under abc, the Catalan power bound from abc.

For the Fermat alternative, the usual arithmetic height statement concerns primitive solutions, or equivalently solutions modulo common scaling. A primitive positive triple is pairwise coprime, and $x,y<z$. Applying abc to $(x^n,y^n,z^n)$ givesThis is uniform even in $n$. If $h=\max (x,y)<z$, then $z^n=x^n+y^n\le 2h^n$, so $n\le \log 2/\log (z/h)$ is bounded after $z$ is bounded. There are therefore finitely many primitive triples and exponents, the primitive Fermat bound from abc. Without the primitive convention, any one nonzero homogeneous solution would produce infinitely many common multiples. The Catalan proof above supplies the requested literal finiteness alternative without that normalization issue.