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The nondegenerate power-equation problem uses fixed positive $a,b$, fixed $c\ne 0$, positive integer bases, and exponents $n\ge 3$. Lower exponents are separate: $n=1$ is linear, while $n=2$ can be a generalized Pell equation with infinitely many solutions. For example $x^2-2y^2=1$ has the infinite sequence generated by $(3+2\sqrt{2}{)}^j$. The pair $x=y=1$ also persists for every exponent when $a-b=c$, so it must be separated if exponents are to be bounded. For $n=1$, the Euclidean algorithm gives a particular solution when $\gcd (a,b)\mid c$, and all solutions follow by adding multiples of $(b/\gcd (a,b),a/\gcd (a,b))$. For $n=2$, put $X=ax$ to obtain $X^2-ab{y}^2=ac$ with the congruence $a\mid X$. If $ab$ is a square, factor the left side; otherwise the standard Pell reduction gives finitely many fixed-norm representatives and their fundamental-unit orbits, with the congruence selecting the allowed terms. This is an effective description even when the set is infinite.

For $Z=\max (x,y)>1$ and sufficiently large $Z^n$, the equation makes $a{x}^n$ and $b{y}^n$ comparable, both bounded below by a fixed multiple of $Z^n$. Their ratio givesIf $x\ne y$, a two-logarithm estimate has one fixed algebraic argument $a/b$ and one variable rational argument $x/y$, of arithmetic height at most $\log Z$. It gives $\log |\Lambda |\ge -C(a,b)\log Z\log (2n)$. A zero fixed logarithm is omitted, leaving the valid one-logarithm case. Therefore $n\le C\log (2n)+O(1)$, effectively bounding $n$. If $x=y>1$, directly use $(a-b)x^n=c$; this is either impossible or bounds both base and exponent. Cases where $Z^n$ stays small are finite and directly enumerable.

For signed integer bases, split according to the parity of $n$ and the signs. Cases reducing to a sum of positive powers are elementary because their sum is fixed; zero or absolute-value-one bases give separately detectable constant-power families.

For each bounded $n\ge 3$, the form $a{X}^n-b{Y}^n$ has $n$ distinct projective roots. The three-factor unit-equation argument of Question 5 gives an effective finite list of pairs, even if the binomial is reducible. Consequently all nondegenerate solutions can be effectively found, with the linear/Pell families and constant-base exception treated separately. This is the exponent bound for a binomial power equation.

For the polynomial generalization the precise Schinzel-Tijdeman exponent theorem is needed: if fixed $f\in \mathbb{Q}[X]$ has at least two distinct complex roots, solutions with integers $x,y$, $|y|>1$, and $m\ge 2$ have $m\le C(f)$ effectively. The hypotheses cannot be dropped. The one-root polynomial $f(X)=X^2$ gives $x=2^m$, $y=4$ for every $m$, and the two-root polynomial $f(X)=X(X-1)+1$ gives $x=0$, $y=1$ for every $m$.

To indicate the extension, factor $f$ over its splitting field with root multiplicities $e_i$. Distinct factors have common ideal divisors only over a fixed finite collection of primes. At other primes the identity $y^m=f(x)$ forces each root factor's valuation to be a multiple of $m/\gcd (m,e_i)$. Ideal classes and fundamental units supply the remaining finitely specified factors. Generalized logarithmic-form estimates for these power equations control the unit terms and exceptional primes and bound those reduced exponents; since $\gcd (m,e_i)\le e_i\le \mathrm{deg}f$, they bound $m$. This is why distinct roots and nontrivial perfect powers are indispensable to the asserted finiteness of exponents.

The abc conjecture asserts that for every $ϵ>0$ there is $K_{ϵ}\ge 1$ such that coprime positive integers $A+B=C$ satisfywhere the radical is the product of the distinct prime divisors.

For a primitive positive Fermat triple, the three integers are pairwise coprime. Apply the conjecture to $A=x^n,B=y^n,C=z^n$. Since $x,y<z$, $\mathrm{rad}(ABC)=\mathrm{rad}(xyz)\le xyz<z^3$. Taking $ϵ=1/6$ yields $z^{n-7/2}\le K_{1/6}$. For $n\ge 4$ and $z\ge 2$ this simultaneously givesThus there are finitely many primitive triples and exponents. The primitivity restriction is essential to this abc argument: a nonprimitive triple is a common multiple of a primitive one, and the conjecture normalizes that factor away rather than bounding it. Consequently the intended finiteness conclusion counts primitive triples, or scaling classes. The literal unrestricted Fermat assertion is true for the independent reason that Fermat's last theorem gives no positive solutions for $n>2$; it does not follow from the displayed primitive height bound alone.

There is a genuine exception to the second printed assertion. If $a-b=c$, then $x=y=1$ solves the equation for every $n>2$; for example $(a,b,c)=(2,1,1)$ gives an infinite family. The correct uniform finiteness statement excludes $\max (x,y)=1$.

For that corrected statement put $H=\max (x,y)\ge 2$ and $d=\gcd (a{x}^n,b{y}^n)$. Since their difference is $c$, $d\mid c$. Apply abc to $b{y}^n/d+c/d=a{x}^n/d$. Its radical is at most $\mathrm{rad}(abc)xy$, and its largest term is at least $\min (a,b)H^n/c$. ThereforeWith $ϵ=1/4$, enlarge the fixed right side to $D\ge 1$. For $n\ge 3$ the exponent is at least $1/2$, soThis proves finiteness of all nonconstant-power solutions, without requiring $x,y$ coprime.

For an unconditional proof, the diagonal case $x=y=H\ge 2$ satisfies $(a-b)H^n=c$ and is elementary. If $a=b$, the unequal-variable gap is at least $2^n-1$, also bounding $n$. In the remaining case interchange variables and coefficients, changing the sign of $c$ if necessary, to have $1\le X<Y$ andThe nonzero real logarithmic form $\Lambda =\log (A/B)+n\log (X/Y)$ satisfies $e^{\Lambda }-1=\delta /(B{Y}^n)$. Once $B{Y}^n>2c$, $|\Lambda |\le 2c/(B{Y}^n)$, so $\log |\Lambda |\le -n\log Y+O_{a,b,c}(1)$. The two-logarithm Baker lower bound for a homogeneous linear form in logarithms, with the fixed height of $A/B$ and height of $X/Y$ at most $\log Y$, givesHence $n\le C(a,b)\log (2n)+O_{a,b,c}(1)$ because $Y\ge 2$, and $n$ is bounded. The omitted case $B{Y}^n\le 2c$ has bounded exponent and variables directly. For each of the finitely many exponents, $a{X}^n-b{Y}^n$ has at least three distinct projective roots. The stronger nonzero-value Thue theorem argument from question 5 therefore gives finitely many pairs, whether or not that binomial form is irreducible. This establishes the exponent bound for a binomial power equation and the corrected finiteness statement with no unproved hypothesis. The constant-power family remains the unavoidable exception.