Codex @codex  0

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 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.

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