Diophantine equation

A Diophantine equation is a polynomial equation for which integer solutions are sought. Congruences, factorization, and size bounds often prove that no solution exists.

A primitive Diophantine solution is an integer tuple satisfying the equation whose coordinates have greatest common divisor one. For homogeneous equations, this removes common integer scale factors.

A triple of positive integers $(a,b,c)$ satisfying $\sqrt[3]{a+b\sqrt{c}}+\sqrt[3]{a-b\sqrt{c}}=1$, with real cube roots. Cubing the equality yields the equivalent integer identity $27b^{2}c=(a+1{)}^2(8a-1)$. In particular $a=3k-1$ for a positive integer $k$, and the condition becomes $b^{2}c=k^2(8k-3)$.

For a fixed polynomial $f\in \mathbb{Q}[X]$ with at least two distinct complex roots, integer solutions with $m\ge 2$ and $|y|>1$ have an effectively bounded exponent. Factorization over a splitting field and the fixed differences between distinct roots control common ideal divisors. Outside a fixed set of primes, each factor's valuation is divisible by $m/\gcd (m,e_i)$, where $e_i$ is its root multiplicity. The generalized power-equation estimates from linear forms in logarithms of algebraic numbers, including unit and exceptional-prime factors, give the exponent bound. A single-root polynomial and the values $y=0,\pm 1$ cannot be included: $4^m=(2^m{)}^2$, and $1^m=f(0)$ for $f(X)=X(X-1)+1$, give infinite exponent families.

A Diophantine equation is soluble at a place of a number field if it has a solution over the corresponding completion of a valued field. It is locally soluble everywhere if this holds at all finite and infinite places. Over the rational numbers, these are the real field and all p-adic fields. Local solubility is necessary for a rational point, but the local-to-global gap in isogeny descent shows it need not suffice.

There are no positive integer solutions to the displayed equation when the exponent is an integer greater than two. Homogeneity means that any hypothetical solution would give infinitely many common multiples; statements bounding solutions by arithmetic height must therefore first distinguish primitive solutions from their multiples.

For every $ϵ>0$ there is a constant $K_{ϵ}$ such that the displayed inequality holds for coprime positive integers $A+B=C$. The radical is the product of the distinct primes dividing the argument. Applying it to a primitive Fermat triple bounds the exponent and the triple simultaneously. For $a{x}^n-b{y}^n=c\ne 0$, division by the gcd, which divides $c$, gives a uniform bound on $\max (x,y{)}^{n-2(1+ϵ)}$. Finiteness in the varying exponent requires excluding $x=y=1$ when $a-b=c$.

A triple of positive integers $(a,b,c)$ with $a<b$, $a+b=c$, $\gcd (a,b)=1$, and $\mathrm{rad}(abc)<c$, where $\mathrm{rad}$ is the radical of an integer. The example $(1,8,9)$ has $\mathrm{rad}(72)=6<9$.

Put $M=x^p$ for positive bases $x,y\ge 2$. The coprime triple $(y^q,1,x^p)$ has radical at most $xy<M^{2/3}$. The abc conjecture with $ϵ=1/6$ gives $M\le C_{1/6}{M}^{7/9}$, proving the bound. Both bases and both exponents are then bounded, since $2^p\le M$ and $2^q\le M-1$. Unlike a homogeneous Fermat equation, no common-scaling ambiguity occurs.

For a primitive positive solution of $x^n+y^n=z^n$ with $n\ge 4$, the three powers are pairwise coprime and their radical is at most $xyz<z^3$. The abc conjecture with $ϵ=1/6$ gives the displayed inequality, hence $z\le C_{1/6}^2$ uniformly in $n$. The inequality $z^n\le 2\max (x,y{)}^n$ then bounds $n$. Thus there are finitely many primitive triples and exponents. Homogeneity would produce infinitely many unscaled solutions from any one nonzero solution, so a primitive arithmetic height bound alone must not be described as finite many arbitrary multiples.

An irreducible integral binary form of degree at least three has only finitely many integral solutions at a fixed nonzero value. Factor the form in a splitting field. Each linear factor generates an ideal dividing a fixed ideal, so it is one of finitely many chosen factors times a unit. The linear relation among any three distinct factors gives a unit equation over a number field, hence finitely many ratios and finitely many rational directions. Homogeneity and the fixed nonzero value then bound the scale on each direction. The same argument works for a binary form with at least three distinct projective roots, even without irreducibility.

For fixed nonzero $\alpha ,\beta$ in a number field, this equation has finitely many unit solutions. By the Dirichlet unit theorem, a large unit-exponent vector makes one unit exponentially small at some archimedean embedding. The equation makes the other fixed multiple exponentially close to one. Its principal logarithm is a nonzero linear form in fixed algebraic logarithms, with coefficients bounded by the exponent size; a polynomial lower bound contradicts the exponential upper bound. Finite torsion and finitely many embeddings allow uniform constants.

An infinite descent rules out a nonempty set of positive integer solutions by constructing from any solution another with a strictly smaller positive integer measure. Choosing a solution of minimal measure gives a contradiction. Common measures are a hypotenuse, a denominator, or the size of a primitive solution.

Choose a primitive counterexample with minimal $z$ and $y$ even. Its Pythagorean triple $(x^2,y^2,z)$ has $x^2=m^2-n^2$, $y^2=2mn$. Modulo four forces $m$ odd and $n$ even, so $m=u^2$ and $n=2v^2$. The primitive triangle $(x,2v^2,u^2)$ then has a parametrization with $v^2=rs$ and $u^2=r^2+s^2$. Coprimality makes $r=a^2,s=b^2$, giving $a^4+b^4=u^2$ with $u<z$. This contradicts infinite descent.