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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.

Diophantus studied algebraic problems seeking numerical solutions, preserved in the Arithmetica. Many of those problems ask for positive rational numbers; their scope therefore differs from the exclusively integer formulation often associated with modern Diophantine equations.^[ref]

Book II, Problem 8 asks for a square to be split into two squares. In modern notation, $16=(12/5{)}^2+(16/5{)}^2$ gives a concrete rational example. The problem connects algebraic manipulation with the search for rational points on a curve, through exact algebraic relations.^[ref]

F. J. M. Barning gave a matrix-based generation process for primitive Pythagorean triples. His 1963 Mathematisch Centrum report verifies the initials and presents three unimodular matrices generating all such triples without repetition, with a fixed leg-order convention.^[ref]

The transformations preserve the equation $x^2+y^2=z^2$. Starting from $(4,3,5)$, with the even leg first, repeated transformations build a branching tree of solutions. The method turns a Diophantine equation into a recursive enumeration algorithm with a uniqueness theorem.^[ref]

A height function measures arithmetic complexity rather than physical size. For a reduced rational number $a/b$, its multiplicative height is $H(a/b)=\max (|a|,|b|)$, and its logarithmic height is $h(a/b)=\log H(a/b)$. Thus $1001/1000$ is close to one in real size but has height $1001$.

For an algebraic number $\alpha$ of degree $d$, let its primitive irreducible integer polynomial have leading coefficient $a_d$ and complex roots ${\alpha }_1,\dots ,{\alpha }_d$. The Absolute logarithmic Weil height is $d^{-1}(\log |a_d|+{\sum }_i\log \max (1,|{\alpha }_i|))$. Bounding both height and degree leaves finitely many possibilities, making heights useful in Diophantine equations and finiteness arguments.

Jan-Hendrik Evertse works in number theory, particularly Diophantine equations. His books study the unit equation over a number field, equations involving discriminants, and effective methods for Diophantine equations over finitely generated integral domains. The effective viewpoint asks for bounds and procedures that make solution sets accessible to explicit computation, beyond an abstract existence or finiteness statement.^[ref]

Homepage: pub.math.leidenuniv.nl/~evertsejh/

Every nonnegative integer is a sum of squares with four terms, with zero terms allowed. Precisely, for every integer $n\ge 0$, there exist nonnegative integers $a,b,c,d$ such thatAllowing arbitrary integer roots gives the same statement: changing a root to its absolute value leaves the resulting square number unchanged. Four is a uniform upper bound on the number of nonzero terms, rather than a demand for four positive terms. Repeated square numbers are allowed. Conrad states the positive case and explicitly explains the role of zero terms; $n=0$ follows by taking all four roots zero. ^[ref]

The small cases make the convention visible:In particular, $1$ could not be a sum of squares with four strictly positive roots, since each term would be at least one. Geometrically, a square of integer side length $a$ has area $a^2$: the theorem says that any nonnegative whole-number total area can be split into at most four such square areas. It does not assert that those squares tile one larger square or a prescribed rectangle.

Why does the fourth term matter? Two square numbers do not suffice for $3$: in modular arithmetic, a square number has remainder zero or one modulo four, so their sum cannot be three modulo four. The sum of two squares theorem gives a complete criterion for the positive integers representable with two terms. Even three terms leave gaps. A square number has remainder $0,1,$ or $4$ modulo eight; adding three of these, with repetition allowed, never gives seven modulo eight. Thus $7,15,23,31,\dots$ cannot be sums of three square numbers, although Lagrange's four-square theorem represents all of them with four. This also proves that the bound four cannot be reduced to three.

The modulo-eight obstruction is not the only way that three terms can fail. For example, $28$ needs four: if three square numbers added to $28$, their remainders modulo four would force all three roots even. Dividing by four would then express $7$ as a sum of three square numbers, contradicting the preceding calculation. These short checks explain the examples without requiring a full classification of sums of three square numbers.

The practical benefit of Lagrange's four-square theorem is an existence guarantee for a family of Diophantine equations: for every nonnegative integer $n$, the displayed equation has an integer solution. It does not specify a unique solution. For instance, $65=8^2+1^2+0^2+0^2=7^2+4^2+0^2+0^2=6^2+4^2+3^2+2^2$. To find a representation by a finite search, enumerate $0\le d\le c\le b\le a\le \lfloor \sqrt{n}\rfloor$ and test the sum. The floor function supplies a finite bound because no single term can exceed $n$, and sorting roots removes duplicate orderings without discarding any possible sum. The theorem guarantees that the search succeeds.

A greedy algorithm that repeatedly subtracts the largest available square number need not find four terms: for $23$ it chooses $16+4+1+1+1$, while $23=9+9+4+1$ uses four. The appeal of the theorem is this gap between local choices and a global promise: individual square numbers become farther apart, and two or three terms leave arithmetic obstructions, yet four terms cover every nonnegative integer.

For historical orientation, Conrad dates Lagrange's four-square theorem to $1770$ and identifies the memoir as Démonstration d'un théorème d'arithmétique: it appeared in the Berlin Academy's volume for the year $1770$, published in $1772$, pages $123$–$133$. Keeping the volume year and publication year distinct avoids an apparent conflict between dates. ^[ref]

A linear form in logarithms of algebraic numbers combines chosen complex logarithms of nonzero algebraic numbers:The coefficients ${\beta }_j$ may be algebraic numbers; the homogeneous integer-coefficient case has ${\beta }_0=0$ and ${\beta }_j=b_j\in \mathbb{Z}$. Each logarithm branch must be fixed, because changing it adds a multiple of $2\pi i$.

When all ${\alpha }_j$ are positive real and real logarithms are used, the homogeneous form is the logarithm of ${\prod }_j{\alpha }_j^{b_j}$. Thus a small nonzero linear form in logarithms of algebraic numbers expresses an unexpectedly close multiplicative relation. For instance,Baker's theorem proves qualitative nonvanishing and transcendence statements, while the general lower bound for a linear form in logarithms gives quantitative separation from zero. Comparing such a lower bound with a small upper bound supplied by a Diophantine equation can bound its exponents, as in the exponent bound for a binomial power equation.

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.

Michel Waldschmidt works on transcendental number theory and Diophantine approximation. With Patrice Philippon, he wrote quantitative results for linear forms in logarithms of algebraic numbers of algebraic numbers.^[ref]

A nonzero expression such as $b_1\log {\alpha }_1+b_2\log {\alpha }_2$ can be small, but lower bounds limit how small it can be in terms of the algebraic data and coefficient sizes. Such bounds convert qualitative nonvanishing into numerical restrictions useful in solving Diophantine equations. The algebraic-coefficient and integer-coefficient cases require distinct estimates.^[ref]

Modular arithmetic performs addition and multiplication on integers while remembering only their remainders for a fixed positive integer modulus $m$. The relation $a\equiv b(\mathrm{mod}m)$ means that $m$ divides $a-b$: the two integers lie in the same congruence class. For example, $-3$, $2$ and $7$ represent one class modulo $5$. Choosing the least nonnegative residue gives exactly $m$ representatives, $0,1,\dots ,m-1$, even though each congruence class contains infinitely many integers.

A clock supplies a familiar example. Represent the twelve positions by $0,1,\dots ,11$, with $0$ at the position usually labelled twelve. Starting at ten and moving seven positions givesThe integer congruence records the final position; it discards the number of complete revolutions. Adding another twelve changes the elapsed count while preserving the position. The picture also shows a single congruence class modulo five spread along the number line.

The useful surprise is that discarding this information still preserves the arithmetic. If $a\equiv a^{\prime }$ and $b\equiv b^{\prime }$ modulo $m$, then $a+b\equiv a^{\prime }+b^{\prime }$ and $ab\equiv a^{\prime }{b}^{\prime }$ modulo $m$. Thus a large calculation can reduce its intermediate integers after every step. For instance,Replacing $10$ by $-2$ gives the same answers with smaller intermediate values. The congruence classes with these operations form the quotient ring $\mathbb{Z}/m\mathbb{Z}$; their addition alone forms a cyclic group. This passage from integers to a finite arithmetic system is the setting for modular exponentiation, primality testing and the RSA cryptosystem. ^[ref]

Multiplicative cancellation needs extra care. Modulo six, $2\cdot 1\equiv 2\cdot 4$, but $1\not\equiv 4$: multiplication by two has merged distinct congruence classes. Cancelling a factor while retaining the integer modulus is justified when that factor has a modular multiplicative inverse, equivalently when it is coprime to the integer modulus. A linear congruence can consequently have no solutions, one solution class, or several solution classes. The Chinese remainder theorem supplies a complementary operation: splitting a calculation across coprime integer moduli and reconstructing its class modulo their product.

Small remainders can also certify impossibility. A perfect square has residue zero or one modulo four, as checking the four congruence classes shows. Therefore an integer equation $x^2=4k+2$ has no solution. The converse direction needs care: a permitted residue does not guarantee an integer solution, since $5\equiv 1(\mathrm{mod}4)$ but five is not a perfect square. Such local tests provide necessary conditions for Diophantine equations without replacing the full problem.

In Disquisitiones arithmeticae (1801), Carl Friedrich Gauss defines integer congruence in article 1 and adopts the symbol $\equiv$ in article 2. Article 12 derives divisibility tests from powers of ten and explains checking arithmetic by its remainders. These passages show a systematic treatment of the same idea; they are not a claim that remainders were first used in 1801. ^[ref]

The Diophantine equation $x^2-D{y}^2=-1$, usually with a nonsquare positive integer $D$ and integer unknowns. It need not have a solution. For $D=2$, multiplying $u+v\sqrt{2}$ by $3+2\sqrt{2}$ preserves norm $-1$ and gives $(u,v)\mapsto (3u+4v,2u+3v)$.

Since $5y^2\le 9$, we have $y\in \{-1,0,1\}$. For $y=0$, $x^2=9$; for $y=\pm 1$, $x^2=4$. Thus the solutions of the Diophantine equation arewhere the two signs in the second family are independent.

The algebraic norm on $\mathbb{Z}[\sqrt{-5}]$ is $N(a+b\sqrt{-5})=a^2+5b^2$ and satisfies multiplicativity of the algebraic norm. The calculation in part (a) therefore gives all elements of norm $9$:If one of these factored into two nonunits, multiplicativity would force both factors to have norm $3$. But the Diophantine equation $a^2+5b^2=3$ has no solution. Hence all six displayed elements are irreducible. This is a standard example in which an irreducible element need not have prime norm.

Write $R=\mathbb{Z}[\sqrt{-13}]$ and $I=(2,1+\sqrt{-13})$. The mapis a surjective ring homomorphism: in ${\mathbb{F}}_2$, the image of $\sqrt{-13}$ is $1$ and $1^2=-13=1$. Its kernel is exactly $I$. Indeed, both generators lie in the kernel; conversely, if $a+b$ is even, thenand $a-b$ is even. Hence $R/I\cong {\mathbb{F}}_2$ and the index of a subgroup is $[R:I]=2$.

If $I=(\alpha )$ were a principal ideal, multiplication by $\alpha =a+b\sqrt{-13}$ would identify its additive lattice with a sublattice of indexThus principality would require $a^2+13b^2=2$. This Diophantine equation has no integer solution: $b=0$ would give $a^2=2$, while $b\ne 0$ makes the left side at least $13$. Therefore

Apply the Euclidean algorithm to $19$ and $11$:Back-substitution gives the Bézout's identityThus $(x,y)=(-4,7)$ is one solution of the Diophantine equation. Subtracting this solution from any other givesSince $19$ and $11$ are coprime, $11$ divides $x+4$. To justify that step directly, multiply $1=-4\cdot 19+7\cdot 11$ by $x+4$: both terms on the right are divisible by $11$, because $19(x+4)=-11(y-7)$. Write $x+4=11t$; substitution then gives $y-7=-19t$. Therefore the complete solution set isEvery such pair satisfies the original Diophantine equation, since the $t$ terms cancel.

Patrice Philippon works on transcendental number theory. His joint chapter with Michel Waldschmidt gives lower bounds for linear forms in logarithms of algebraic numbers, explicitly distinguishing general algebraic coefficients from rational-integer coefficients.^[ref]

The central arithmetic question is quantitative: once a logarithmic combination is known to be nonzero, how closely can it approach zero? An explicit bound links Diophantine approximation to finite calculations and can help restrict candidate solutions of Diophantine equations.

A quadratic integer is an algebraic integer in a quadratic number field $K=\mathbb{Q}(\sqrt{d})$, where $d$ is a squarefree integer other than $0,1$. Its ring of integers is $\mathbb{Z}[(1+\sqrt{d})/2]$ if $d\equiv 1(\mathrm{mod}4)$, and $\mathbb{Z}[\sqrt{d}]$ otherwise. Thus $(1+\sqrt{5})/2$ is a quadratic integer, since it satisfies the monic equation $X^2-X-1=0$.

For $\alpha =a+b\sqrt{d}$, its conjugate is $a-b\sqrt{d}$. Its field trace and field norm are $2a$ and $a^2-d{b}^2$; their integrality characterizes the algebraic integers in this degree-two extension. These formulas connect Diophantine equations to arithmetic in a larger ring.