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For fixed positive $a,b$ and fixed nonzero $c$, with $x\ne y$, put $Y=\max (x,y)$ and take $\Lambda =\log (a/b)+n\log (x/y)$, swapping the variables and coefficients if necessary. The equation gives $\log |\Lambda |\le -n\log Y+O(1)$ when the larger-variable power is sufficiently large. A two-logarithm lower bound gives $\log |\Lambda |\ge -C(a,b)\log Y\log (2n)$. Hence $n\le C\log (2n)+O(1)$, so $n$ is bounded. The cases $x=y$ and a bounded larger-variable power are elementary. For each remaining exponent, the Thue theorem for a binomial form supplies finitely many pairs. The constant-power exception $x=y=1$, $a-b=c$, must be removed.

Fix fundamental units ${\epsilon }_1,\dots ,{\epsilon }_r$ and write $x=\zeta \prod {\epsilon }_i^{b_i}$. Rank zero already gives a finite unit group. Otherwise put $B=\max (1,|b_i|)$. The logarithmic embedding in Dirichlet's theorem has a weighted sum-zero image and is injective on the free exponent lattice. Norm equivalence and that sum-zero condition give an archimedean embedding $\sigma$ withThis is the small archimedean value of a large unit. From the equation, $\sigma (\beta y)=1-\sigma (\alpha x)$ is exponentially close to one. Also $h(y)\le h(x)+O(1)$ by the arithmetic height inequalities for $y=(1-\alpha x)/\beta$, so the fundamental-unit exponents of $y$ have size $O(B)$.

For large $B$, take the small-branch logarithm of $\sigma (\beta y)$. With $y=\eta \prod {\epsilon }_i^{c_i}$ it is the nonzero formusing fixed logarithms and $\log (-1)=\pi i$. The integer $k$ corrects the branch, and nonzero follows since $\beta y=1$ would force $\alpha x=0$. Its upper bound is $|\Lambda |\le C^{\prime }{e}^{-cB}$. The assumed effective logarithmic-form estimate for the fixed algebraic numbers gives $\log |\Lambda |\ge -C^{\prime \prime }\log (2B)$. Thus $cB\le C^{\prime \prime }\log (2B)+O(1)$, effectively bounding $B$. The finitely many embeddings and torsion choices give uniform constants. Therefore there are only finitely many unit solutions, and their exponents are effectively bounded.

For the Thue application, take an irreducible integral binary form $F$ of degree $d\ge 3$ and $m\ne 0$. In a splitting field, write $F(X,Y)=a{\prod }_i(X-{\rho }_{i}Y)$ and put ${\theta }_i=a{\rho }_i$, which are algebraic integers. The integral factors $L_i=aX-{\theta }_{i}Y$ satisfy ${\prod }_i{L}_i=a^{d-1}m$. Each ideal $(L_i)$ therefore divides one fixed ideal, giving finitely many possibilities. Choose a generator ${\delta }_i$ for each principal possibility, so $L_i={\delta }_i{u}_i$ with units $u_i$.

For three distinct roots,Division by the third term gives a fixed-coefficient unit equation in $u_1/u_3,u_2/u_3$. Its solutions are finite, hence there are finitely many values of $L_1/L_2$. This ratio determines the rational direction $X:Y$, since the roots are distinct. In a fixed direction, homogeneity and $F(X,Y)=m\ne 0$ leave only finitely many integer scales. This proves Thue theorem, and the argument also works for a form with at least three distinct projective roots even if reducible.

The hyperelliptic alternative uses a related factorization for the usual nondegenerate case with squarefree $f$ of degree at least three. For $y^2=f(x)$, ideals of two different root factors can share primes only above the fixed root differences and the leading coefficient. At other primes their valuations are even. Finite exceptional-prime and ideal-class choices, together with units modulo squares, reduce the factors to finitely many expressions ${\delta }_i{z}_i^2$. Relations between distinct root factors then give simultaneous norm or Pell-type equations. Writing their solutions in fundamental units and applying the same small-value/logarithmic-form estimates bounds the unit exponents, hence $x,y$. The common-factor and square-class reductions are essential; a single quadratic norm equation alone can have infinitely many Pell solutions. This outlines the effective hyperelliptic treatment as well as the explicit Thue reduction.

Use the following fixed-number form of a logarithmic estimate: for chosen logarithms of fixed nonzero algebraic numbers $a_1,\dots ,a_s$, a nonzero form $\Lambda =\sum b_j\log a_j$ with integer coefficients $|b_j|\le B$ satisfies $|\Lambda |\ge c(2B{)}^{-C}$, for fixed positive constants $c,C$. This follows from the Baker lower bound for a homogeneous linear form in logarithms.

By the Dirichlet unit theorem, write units as $x=\zeta \prod {ϵ}_j^{m_j}$ and $y=\xi \prod {ϵ}_j^{n_j}$, with finitely many roots of unity and fixed fundamental units. If the unit rank is zero, there are already only finitely many choices. Otherwise put $B=\max (1,|m_j|,|n_j|)$. The real logarithmic embedding of the free unit group has full rank and lies in the product-formula hyperplane. Equivalence of norms in that finite-dimensional space shows that a unit with exponent size $B$ has some archimedean conjugate of modulus at most $e^{-c_{0}B}$: a large positive logarithmic coordinate forces a comparably large negative one, and a large negative coordinate itself suffices.

Interchanging the two terms if necessary, choose such a conjugate for $y$. At this embedding the equation makes $\sigma (\alpha x)=1-\sigma (\beta y)$ with $|\sigma (\beta y)|\le C_0{e}^{-c_{0}B}$. For large $B$ define the logarithm near one. It is nonzero because $\beta y\ne 0$, andOn fixed branches it isfor an integer $k=O(B)$. The bound on $k$ follows from summing the fixed imaginary parts and then reducing to the near-one branch. There are only finitely many possible embeddings and roots of unity, so the stated lower bound applies uniformly with coefficients $O(B)$. It contradicts the exponential upper bound when $B$ is large. Hence the exponents are bounded and $\boxed{\alpha x+\beta y=1\text{ has finitely many unit solutions}}$ in ${\mathcal{O}}_K^{\times }$. This is the unit equation over a number field proof; “unit” here means an algebraic-integer unit, not an arbitrary nonzero field element.

For the Thue theorem, let $F\in \mathbb{Z}[X,Y]$ be irreducible homogeneous of degree $d\ge 3$ and let $m\ne 0$ be fixed. In a splitting field $L$, write $F(X,Y)=a{\prod }_i(X-{\theta }_{i}Y)$ with distinct roots. Choose a fixed integer $c$ so that each $c{\theta }_i$ is integral and $c^{d}m/a$ is a nonzero integer. For a solution $(u,v)$ the nonzero algebraic integers $t_i=c(u-{\theta }_{i}v)$ satisfy ${\prod }_i{t}_i=c^{d}m/a$. Their principal ideals therefore divide a fixed ideal and have only finitely many possibilities. For each occurring principal ideal choose a generator ${\gamma }_i$, so $t_i={\gamma }_i{\epsilon }_i$ with ${\epsilon }_i$ a unit.

Any three distinct roots give the identityDivide by the third term and absorb the finitely many generator ratios into the coefficients. This is a fixed-coefficient unit equation in ${\epsilon }_1/{\epsilon }_3$ and ${\epsilon }_2/{\epsilon }_3$, so it has finitely many solutions for each generator choice. Thus $\rho =t_1/t_3$ has finitely many values. The relation $(1-\rho )u=({\theta }_1-\rho {\theta }_3)v$ fixes a rational direction whenever an integer solution exists; $\rho =1$ simply gives $v=0$. On each such direction, $(u,v)=k(u_0,v_0)$ for a primitive integer pair, and $k^{d}F(u_0,v_0)=m$ allows finitely many $k$. This proves the desired finiteness. When $m=0$, irreducibility excludes a nonzero rational root direction, leaving only $(0,0)$.

The argument also proves finiteness for any binary form with at least three distinct projective roots: irreducibility was used only to guarantee distinct roots and to dispose of the zero-value case. A rational invertible change of variables can move every projective root away from infinity; after clearing its fixed denominators, the resulting pairs lie in an integer lattice and the same proof applies. This stronger nonzero-value version will apply to the binomial forms in question 6.

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.