Diophantine equation

A Diophantine equation is a polynomial equation for which integer solutions are sought. Here the polynomial coefficients are also integers. A real solution alone need not meet this arithmetic requirement.

For example, $x^2+y^2=z^2$ has the Pythagorean triple $(3,4,5)$ as a solution. In contrast, $x^2+y^2=3$ has no integer solution: a perfect square is zero or one modulo four, so their sum cannot be three. The circle still has real points, such as $(\sqrt{3},0)$.

A linear Diophantine equation is controlled by a greatest common divisor and Bézout's identity. Higher-degree equations can require integer factorization, quadratic reciprocity, infinite descent or bounds on arithmetic height. A primitive Diophantine solution separates common scaling when the equation is defined by a homogeneous polynomial, while local solubility tests the equation over completions before attempting a global solution.

In the ring of integers $\mathbb{Z}[\sqrt{-6}]$, which has class number two and units $\pm 1$, write $\alpha =x+(y+1)\sqrt{-6}$. Its norm is $(y+2{)}^3$. Parity modulo eight and cubic residues modulo nine show that $y$ is odd and $3$ does not divide $y+2$, so $\gcd (y+2,6)=1$. The ideals of $\alpha$ and its conjugate are coprime: a common prime cannot lie above two or three by their norms; at any other prime their difference forces divisibility of $y+1$, contradicting divisibility of $y+2$.

Unique factorization of ideals gives $(\alpha )=A^3$. The class group of order two forces $A$ principal, and its unit factor can be absorbed into a cube. Hence $\alpha =(u+v\sqrt{-6}{)}^3$ with integers $u,v$, and $u^2+6v^2=y+2$. Comparing coefficients and eliminating $y$ yieldsIntegrality requires $3v-1=\pm 1$, so $v=0$ is the only integral possibility. Then $u=\pm 1$, $y=-1$, and $x=\pm 1$. Direct substitution verifies both solutions. The class-group step is what permits extracting a cube of an element from a cube of an ideal despite failure of unique factorization of elements.

The equation has no integer solutions. Work in the ring of integers $\mathbb{Z}[\theta ]$ for $\theta =\sqrt{-30}$. If a rational prime divides both $y$ and thirty, it divides $x$, and the valuation of $x^2+30$ is one because thirty is squarefree, contradicting the valuation of a cube. Thus $\gcd (y,30)=1$. The two principal ideals $(x+\theta ),(x-\theta )$ are coprime: any common prime ideal lies above a prime dividing both $N(2\theta )=120$ and $y$. Their product is $(y{)}^3$, so unique prime ideal factorization gives $(x+\theta )=A^3$. The Ideal class group of Q of square root minus thirty has exponent two, forcing $A$ to be principal. The only units are $\pm 1$ and both are cubes, so $x+\theta =(a+b\theta {)}^3$ for integers $a,b$. Comparison of the coefficient of $\theta$ gives $1=3b(a^2-10b^2)$, impossible modulo three.

Cornacchia's algorithm searches for primitive integer solutions of $x^2+d{y}^2=m$ by combining a modular square root of $-d$ modulo $m$ with the Euclidean algorithm. In the standard odd-prime case $m=p$, choose a root $r$ with $0<r<p/2$, run the Euclidean algorithm on $(p,r)$ until a remainder $x<\sqrt{p}$ occurs, and test whether $(p-x^2)/d$ is a nonnegative integer square. Both roots can be tried.

For $p=13,d=1$, the root $r=5$ of $-1$ gives the remainder $3$, and $13-3^2=4$, yielding $(x,y)=(3,2)$. The final exact square test is essential: a modular root is only a candidate-producing input. For general composite moduli, additional hypotheses and choices are required to enumerate primitive solutions.

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 Cardano triplet is a triple of positive integers $(a,b,c)$ satisfyingwith real cube roots. Put $u=\sqrt[3]{a+b\sqrt{c}}$ and $v=\sqrt[3]{a-b\sqrt{c}}$. Cubing $u+v=1$ gives $uv=(1-2a)/3$, and cubing that product yields the necessary integer identityIt is also sufficient. From this identity, the real cube root of $a^2-b^{2}c$ gives $uv=(1-2a)/3$. Therefore $s=u+v$ satisfiesSince $a\ge 1$, the quadratic polynomial has negative discriminant $1-8a$ and no real roots. Hence $s=1$. This checks that cubing introduced no extraneous real solutions.

Reducing the integer identity modulo three forces $a\equiv 2(\mathrm{mod}3)$. Writing $a=3k-1$ for a positive integer $k$ then reduces it to $b^{2}c=k^2(8k-3)$. For example, $k=1$ gives $(a,b,c)=(2,1,5)$.

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 $ϵ>0$, the abc conjecture asserts that there is a constant $K_{ϵ}>0$ such that all coprime positive integers $A+B=C$ satisfyThe radical of an integer is the product of its distinct prime factors. The condition $\gcd (A,B)=1$ makes the three entries pairwise coprime. Repeated prime powers enlarge a number without enlarging its radical in the same way, so the conjecture constrains how often addition can relate unusually concentrated prime factorizations.

The example $(A,B,C)=(1,8,9)$ has $\mathrm{rad}(ABC)=6<9$ and is an abc-hit. Such examples are compatible with the conjecture: its bound includes both the constant $K_{ϵ}$ and the exponent $1+ϵ$. Conditional applications include the primitive Fermat bound from abc and the Catalan power bound from abc.

More generally, fix positive integers $a,b$, a nonzero integer $c$, and positive integers $x,y,n$ satisfying $a{x}^n-b{y}^n=c$. Put $g=\gcd (a{x}^n,b{y}^n)$, which divides $|c|$. Applying the conjecture to the normalized positive triple formed from $a{x}^n$, $b{y}^n$ and $|c|$ gives, for $M=\max (x,y)$,Indeed, the largest normalized entry is at least a fixed multiple of $M^n$, while the radical is at most $\mathrm{rad}(ab|c|)xy\le \mathrm{rad}(ab|c|)M^2$. Taking $ϵ=1/4$ makes the exponent at least $1/2$ for $n\ge 3$, bounding $M$ uniformly; if $M\ge 2$, it also bounds $n$. The separate case $x=y=1$ can satisfy the equation for every $n$ 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 a number field $K$ and fixed nonzero coefficients $\alpha ,\beta \in K$, the equationhas finitely many solutions in the unit group of its ring of integers. The coefficients remain fixed; varying them gives a different finiteness problem.

Here is the logarithmic-form mechanism. The Dirichlet unit theorem writes each unit as a root of unity times a product of fixed fundamental units with integer exponents. If the unit rank is zero there are already only finitely many possibilities. Otherwise set $B$ to the maximum of $3$ and the absolute values of their exponents. The logarithmic embedding and the product formula imply that a unit with a large exponent vector is exponentially small at some archimedean place, after possibly interchanging $u$ and $v$.

At that place, $\alpha u+\beta v=1$ makes the other term exponentially close to one. A nonzero logarithm of that term is a linear form in logarithms of algebraic numbers, including a suitable multiple of $2\pi i$ when complex logarithm branches are adjusted. Its integer coefficients are $O(B)$. The Baker lower bound for a homogeneous linear form in logarithms bounds its nonzero size below by $B^{-C}$ for fixed data, contradicting an exponentially small upper bound when $B$ grows. The logarithm cannot vanish, since then the other term would be exactly one and the remaining nonzero term zero. Thus the exponent vectors are bounded, leaving finitely many solutions. A finite torsion subgroup and finitely many archimedean places 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.