Past exam of the mathematics course of the University of Cambridge / 2001 / iii / Paper 20 / 6

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 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.