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A figurate number is an integer illustrated by a regular geometric arrangement of points. Triangular numbers and square numbers are basic examples: growing a prescribed shape turns a counting problem into a sequence of integers.

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]

For a set $S$ of nonnegative integers, its natural density is the limiting proportion of the first $N$ nonnegative integers that belong to $S$:provided this limit exists. Write $C_S(N)=|S\cap \{0,\dots ,N-1\}|$ for the counting function. The numerator is a finite cardinality; the denominator is exactly $N$, including zero. Every existing natural density lies between zero and one.

For the even nonnegative integers $E=\{0,2,4,\dots \}$, there are $m$ elements below $N=2m$ and $m+1$ below $N=2m+1$. Thus $C_E(N)=\lceil N/2\rceil$ using the ceiling function, and $d(E)=1/2$. For the square numbers $Q=\{0,1,4,9,\dots \}$, the exact count for $N\ge 1$ isThe floor function counts the allowed indices $1,\dots ,\lfloor \sqrt{N-1}\rfloor$, and the extra one counts zero. At $N=10$, the square numbers below ten are $0,1,4,9$, so the proportion is $4/10$; at $N=100$, it is $10/100$. The count grows like the square root of $N$, while the denominator grows like $N$.

The prime numbers also have natural density zero. If $P$ is their set, the Prime number theorem gives $C_P(N)\sim N/\log N$, hence $C_P(N)/N\to 0$. The Prime number theorem is a substantial result about the count, rather than a consequence of the fact that there are infinitely many prime numbers. ^[ref]

Why use natural density when cardinality already measures size? The even nonnegative integers, the square numbers, and the prime numbers all have the same countably infinite cardinality, yet their proportions among large initial integer intervals differ. Natural density distinguishes frequent membership from sparse membership. Equivalently, under a uniform distribution on a finite set of $N$ possible nonnegative integers, $C_S(N)/N$ is the probability of landing in $S$. A set without natural density shows why this limiting proportion need not exist.

Reduce $z$ using modular congruence: its residue class modulo $3$ is $0$, $1$, or $-1$. A square number therefore has residue class $0$ or $1$, and raising either residue to the positive integer power $n$ leaves it unchanged. ConsequentlyThe exponent in the original PDF is $2n$; the polynomial identities in the converted TeX are a transcription error. The argument uses $n\ge 1$; it does not assign $0^0$ a value.

For a Pythagorean triple, each square number is $0$ or $1$ modulo $3$. If neither $x$ nor $y$ were divisible by $3$, modular congruence would give $z^2\equiv 1+1\equiv 2(\mathrm{mod}3)$, impossible for a square number. Thus at least one of $x,y$ is divisible by $3$; both can be divisible, as in $(9,12,15)$.

An odd number has square number congruent to $1$ modulo $4$, whereas an even number has square number congruent to $0$. If both $x$ and $y$ were odd numbers, their sum of square numbers would again be $2$ modulo $4$, which no square number can be. Hence $x$ and $y$ cannot both be odd. They can both be even; the conclusion does not require the Pythagorean triple to be primitive.

An arithmetic function $f$ is multiplicative if $f(mn)=f(m)f(n)$ whenever $m$ and $n$ are coprime integers. If $g(n)={\sum }_{d\mid n}f(d)$ and $(m,n)=1$, every positive integer divisor of $mn$ is uniquely $d_1{d}_2$ with $d_1\mid m$ and $d_2\mid n$. Henceso $g$ is multiplicative.

The Möbius function is defined by $\mu (1)=1$, $\mu (n)=0$ if the square of a prime number divides $n$, and $\mu (n)=(-1{)}^r$ if $n$ is a product of $r$ distinct prime numbers. Coprime integers have disjoint sets of prime factors, so this definition immediately gives $\mu (mn)=\mu (m)\mu (n)$ when $(m,n)=1$.

For $n=p_1^{a_1}\cdots p_r^{a_r}>1$, only the square-free divisors contribute, and thereforeFor $n=1$ the sum is $1$. Thus $1\ast \mu =\epsilon$ in the language of Dirichlet convolution. Convolving $g=1\ast f$ with $\mu$ gives the Möbius inversion formula

If $f(n)=n$, its divisor sum is the sum-of-divisors function. Conversely, if $g(n)=n$, Möbius inversion givesusing the standard divisor identity for the Euler totient function. Therefore the two requested answers are

LetOver ${\mathbb{Q}}_2$, the Newton polygon has three length-one segments of slopes $-2,-1,0$. The resulting roots can also be obtained directly from Hensel lemma. There is one unit root because $F(1)$ is even and $F^{\prime }(1)$ is odd. For the two remaining roots, putwhose parenthesized polynomial has a simple root $Y\equiv 1(\mathrm{mod}2)$, andwhose parenthesized polynomial has a simple root $Z\equiv 1(\mathrm{mod}2)$. Thus $F$ splits completely over ${\mathbb{Q}}_2$, with roots of valuations $0,1,2$. There are three primes $q$ above $2$, and for each one

Over ${\mathbb{Q}}_5$, $F$ is an Eisenstein polynomial. Hence there is one prime $q$ above $5$, and if $\alpha$ is the chosen root thenwith $e=3$ and $f=1$.

The polynomial is irreducible over $\mathbb{Q}$ by the Eisenstein criterion at $5$. Its polynomial discriminant iswhich is not a square number, so the Galois group of an irreducible cubic isAt $2$, all three roots already lie in ${\mathbb{Q}}_2$, so the local splitting field is trivial andAt $5$, the cubic is totally and tamely ramified. The square class of its discriminant is represented by $-13807\equiv 3(\mathrm{mod}5)$, a nonsquare unit, so the quadratic resolvent field of a cubic is the unramified quadratic extension of ${\mathbb{Q}}_5$. The local splitting field therefore has degree six, withThese calculations are summarized by local factorization of X3 plus 25X2 minus 50X plus 40.

The language $L=\{a^{m^2}{b}^{m^2}:m\ge 0\}$ is not context-free. Suppose it had pumping length $p$ in the pumping lemma for context-free languages, and apply the lemma to $w=a^{p^2}{b}^{p^2}$. Write $w=uvxyz$ with $|vxy|\le p$, $|vy|>0$, and $u{v}^{i}x{y}^{i}z\in L$ for every $i\ge 0$.

Let $\alpha$ and $\beta$ be the respective numbers of $a$s and $b$s in $vy$. If $\alpha \ne \beta$, pumping with $i=0$ or $i=2$ destroys equality of the two block lengths. Hence a valid decomposition would require $\alpha =\beta =d>0$. Pumping with $i=2$ would then produce $a^{p^2+d}{b}^{p^2+d}$, where $1\le d\le p$. Butso $p^2+d$ is not a square number. This contradiction proves the square-count language is not context-free.

For integers $s\ge 3$ and $n\ge 1$, the $n$th $s$-gonal number is $P_{s,n}=((s-2)n^2-(s-4)n)/2$. Its successive differences form an arithmetic progression. The cases $s=3,4,5,6$ give triangular numbers, square numbers, pentagonal numbers, and hexagonal numbers.

A Pythagorean triple consists of three positive integers satisfying $x^2+y^2=z^2$. It is primitive when their greatest common divisor is one. Reducing their square numbers modulo $3$ shows that at least one leg is divisible by $3$; reducing modulo $4$ shows that both legs cannot be odd.

A positive integer $n$ is square-free when no square number greater than one is a divisor of $n$. It suffices to exclude squares of prime numbers; equivalently, every prime valuation is zero or one:Thus the prime factorization of a square-free integer is a product of distinct prime factors, with no repetitions. One is included: its prime factorization is an empty product and all its prime valuations are zero. These equivalent tests are stated in Grinberg's notes. ^[ref]

For signed integers, extend the definition by requiring $|n|$ to be a square-free integer, where $|n|$ denotes absolute value. Hence $-30$ and both $1,-1$ qualify. Zero does not: every prime number has its square as a divisor of zero. The Möbius function and radical of an integer below use their existing positive integer domains, so apply them to $|n|$ when handling a negative integer.

For example, $30=2\cdot 3\cdot 5$ is a square-free integer, whereas $12=2^2\cdot 3$ is not, because $4\mid 12$ and $v_2(12)=2$. Being a composite number causes no problem: $6,10,15$ are all square-free integers. A square-free integer need not be coprime to another integer: $6$ and $10$ are both square-free integers but have greatest common divisor two. In fact the only signed integers coprime to every integer are $1,-1$, since $\gcd (n,n)=|n|$.

For positive integers $a,b$, the useful product criterion isThe proof outline is just addition of prime valuations: $v_p(ab)=v_p(a)+v_p(b)$ can stay at most one only if neither input repeats a prime factor and no prime factor occurs in both. Thus $6\cdot 5=30$ is a square-free integer, but $6\cdot 10=60$ is not. An equivalent characterization is that every factorization $n=ab$ into positive integers has coprime factors. If $p^2\mid n$, the factorization $n=p(n/p)$ supplies two factors sharing $p$; conversely, a shared prime factor forces its square to divide $n$. For $30$, the unordered factor pairs $(1,30),(2,15),(3,10),(5,6)$ all have greatest common divisor one; for $12$, the pair $(2,6)$ fails.

Two existing arithmetic functions turn the definition into practical tests:The radical of an integer keeps one copy of each prime factor, so it changes precisely the integers with a repeated prime factor. Its definition is given in Evertse's exercise sheet. ^[ref] The Möbius function assigns $(-1{)}^r$ to a square-free integer with $r$ distinct prime factors and zero otherwise; squaring forgets the sign. Consequently ${\sum }_{n=1}^{60}\mu (n{)}^2=37$ counts exactly the green cells. This connects a restriction on repeated prime factors with weighted counting and inclusion-exclusion. ^[ref]