Innumber theory, the sum of the firstncubes is thesquare of thenthtriangular number. That is,

${\displaystyle 1^3+2^3+3^3+\cdots +n^3={\left(1+2+3+\cdots +n\right)}^2.}$

The same equation may be written more compactly using the mathematical notation forsummation:

${\displaystyle \sum _{k=1}^n{k}^3={\left(\sum _{k=1}^{n}k\right)}^2.}$

Thisidentity is sometimes calledNicomachus's theorem, afterNicomachus of Gerasa (c. 60 –c. 120 CE).

Nicomachus, at the end of Chapter 20 of hisIntroduction to Arithmetic, pointed out that if one writes a list of the odd numbers, the first is the cube of 1, the sum of the next two is the cube of 2, the sum of the next three is the cube of 3, and so on. He does not go further than this, but from this it follows that the sum of the first${\displaystyle n}$ cubes equals the sum of the first${\displaystyle \frac{n(n+1)}{2}}$ odd numbers, that is, the odd numbers from 1 to${\displaystyle n(n+1)-1}$. The average of these numbers is obviously${\displaystyle \frac{n(n+1)}{2}}$, and there are${\displaystyle \frac{n(n+1)}{2}}$ of them, so their sum is${\displaystyle {\left(\frac{n(n+1)}{2}\right)}^2}$.

Many early mathematicians have studied and provided proofs of Nicomachus's theorem.Stroeker (1995) claims that "every student of number theory surely must have marveled at this miraculous fact".^[1]Pengelley (2002) finds references to the identity not only in the works ofNicomachus in what is nowJordan in the first century CE, but also in those ofAryabhata inIndia in the 5th century, and in those ofAl-Karajic. 1000 inPersia.^[2]Bressoud (2004) mentions several additional early mathematical works on this formula, byAl-Qabisi (10th century Arabia),Gersonides (c. 1300, France), andNilakantha Somayaji (c. 1500, India); he reproduces Nilakantha's visual proof.[3]

The sequence of squared triangular numbers is

These numbers can be viewed asfigurate numbers, a four-dimensional hyperpyramidal generalization of thetriangular numbers andsquare pyramidal numbers.

AsStein (1971) observes, these numbers also count the number of rectangles with horizontal and vertical sides formed in an${\displaystyle n\times n}$grid. For instance, the points of a${\displaystyle 4\times 4}$ grid (or a square made up of three smaller squares on a side) can form 36 different rectangles. The number of squares in a square grid is similarly counted by the square pyramidal numbers.^[4]

The identity also admits a natural probabilistic interpretation as follows. Let${\displaystyle X,Y,Z,W}$ be four integer numbers independently and uniformly chosen at random between 1 and${\displaystyle n}$. Then, the probability that${\displaystyle W}$ is the largest of the four numbers equals the probability that${\displaystyle Y}$ is at least as large as${\displaystyle X}$ and that${\displaystyle W}$ is at least as large as${\displaystyle Z}$. That is,$${\displaystyle Pr[max(X,Y,Z)\le W]=Pr[X\le Y\wedge Z\le W].}$$ For any particular value of${\displaystyle W}$, the combinations of${\displaystyle X}$,${\displaystyle Y}$, and${\displaystyle Z}$ that make${\displaystyle W}$ largest form a cube${\displaystyle 1\le X,Y,Z\le n}$ so (adding the size of this cube over all choices of${\displaystyle W}$}) the number of combinations of${\displaystyle X,Y,Z,W}$ for which${\displaystyle W}$ is largest is a sum of cubes, the left hand side of the Nichomachus identity. The sets of pairs${\displaystyle (X,Y)}$ with${\displaystyle X\le Y}$ and of pairs${\displaystyle (Z,W)}$ with${\displaystyle Z\le W}$ form isosceles right triangles, and the set counted by the right hand side of the equation of probabilities is theCartesian product of these two triangles, so its size is the square of a triangular number on the right hand side of the Nichomachus identity. The probabilities themselves are respectively the left and right sides of the Nichomachus identity, normalized to make probabilities by dividing both sidesby${\displaystyle n^4}$.^{[citation needed]}

Charles Wheatstone (1854) gives a particularly simple derivation, by expanding each cube in the sum into a set of consecutive odd numbers. He begins by giving the identity$${\displaystyle n^3=\underset{n\text{ consecutive odd numbers}}{\underbrace{\left(n^2-n+1\right)+\left(n^2-n+1+2\right)+\left(n^2-n+1+4\right)+\cdots +\left(n^2+n-1\right)}}.}$$That identity is related totriangular numbers${\displaystyle T_n}$ in the following way:$${\displaystyle n^3=\sum _{k=T_{n-1}+1}^{T_n}(2k-1),}$$and thus the summands forming${\displaystyle n^3}$ start off just after those forming all previous values${\displaystyle 1^3}$ up to${\displaystyle (n-1{)}^3}$. Applying this property, along with another well-known identity:$${\displaystyle n^2=\sum _{k=1}^n(2k-1),}$$produces the following derivation:^[5]

Row (1893) obtains another proof by summing the numbers in a squaremultiplication table in two different ways. The sum of theith row isi times a triangular number, from which it follows that the sum of all the rows is the square of a triangular number. Alternatively, one can decompose the table into a sequence of nestedgnomons, each consisting of the products in which the larger of the two terms is some fixed value. The sum within each gnomon is a cube, so the sum of the whole table is a sum of cubes.^[6]

In the more recent mathematical literature,Edmonds (1957) provides a proof usingsummation by parts.^[7]Stein (1971) uses the rectangle-counting interpretation of these numbers to form a geometric proof of the identity.^[8] Stein observes that it may also be proved easily (but uninformatively) by induction, and states thatToeplitz (1963) provides "an interesting old Arabic proof".^[4]Kanim (2004) provides a purely visual proof,^[9]Benjamin & Orrison (2002) provide two additional proofs,^[10] andNelsen (1993) gives seven geometric proofs.^[11]

A similar result to Nicomachus's theorem holds for allpower sums, namely that odd power sums (sums of odd powers) are a polynomial in triangular numbers.These are calledFaulhaber polynomials, of which the sum of cubes is the simplest and most elegant example.However, in no other case is one power sum a square of another.^[7]

Stroeker (1995) studies more general conditions under which the sum of a consecutive sequence of cubes forms a square.^[1]Garrett & Hummel (2004) andWarnaar (2004) study polynomial analogues of the square triangular number formula, in which series of polynomials add to the square of another polynomial.^[12]

• Weisstein, Eric W.,"Nicomachus's theorem",MathWorld
• A visual proof of Nicomachus's theoremArchived 2019-10-19 at theWayback Machine

• Number theory
• Integer sequences
• Algebraic identities
• Proof without words

• Articles with short description
• Short description is different from Wikidata
• All articles with unsourced statements
• Articles with unsourced statements from July 2021
• CS1 maint: ignored ISBN errors
• Webarchive template wayback links
• Articles containing proofs