= Solution
The nondegenerate power-equation problem uses fixed positive $a,b$, fixed $c\ne0$, positive <integer> bases, and exponents $n\ge3$. 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 $(3+2\sqrt2)^j$. 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 $ax^n$ and $by^n$ comparable, both bounded below by a fixed multiple of $Z^n$. Their ratio gives
$$
\Lambda=\log(a/b)+n\log(x/y)=\log(1+c/(by^n))\ne0,\qquad\log|\Lambda|\le-n\log Z+O(1).
$$
If $x\ne y$, 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 $n\ge3$, the form $aX^n-bY^n$ 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 \b[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 $m\ge2$ 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 $\gcd(m,e_i)\le e_i\le\deg f$, they bound $m$. This is why distinct roots and nontrivial perfect powers are indispensable to the asserted finiteness of exponents.
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