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    History of the Theory of Numbers, Volume II: Diophantine Analysis
    History of the Theory of Numbers, Volume II: Diophantine Analysis
    History of the Theory of Numbers, Volume II: Diophantine Analysis
    Ebook1,707 pages37 hours

    History of the Theory of Numbers, Volume II: Diophantine Analysis

    By Leonard Eugene Dickson

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    The three-volume series History of the Theory of Numbers is the work of the distinguished mathematician Leonard Eugene Dickson, who taught at the University of Chicago for four decades and is celebrated for his many contributions to number theory and group theory. This second volume in the series, which is suitable for upper-level undergraduates and graduate students, is devoted to the subject of diophantine analysis. It can be read independently of the preceding volume, which explores divisibility and primality, and volume III, which examines quadratic and higher forms.
    Featured topics include polygonal, pyramidal, and figurate numbers; linear diophantine equations and congruences; partitions; rational right triangles; triangles, quadrilaterals, and tetrahedra; the sums of two, three, four, and n squares; the number of solutions of quadratic congruences in n unknowns; Liouville's series of eighteen articles; the Pell equation; squares in arithmetical or geometrical progression; equations of degrees three, four, and n; sets of integers with equal sums of like powers; Waring's problem and related results; Fermat's last theorem; and many other related subjects. Indexes of authors cited and subjects appear at the end of the book.
    LanguageEnglish
    PublisherDover Publications
    Release dateOct 17, 2013
    ISBN9780486154602
    History of the Theory of Numbers, Volume II: Diophantine Analysis

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      History of the Theory of Numbers, Volume II - Leonard Eugene Dickson

      INDEX

      CHAPTER I.

      POLYGONAL, PYRAMIDAL AND FIGURATE NUMBERS.

      The formation of triangular numbers 1, 1 + 2, 1 + 2 + 3, …, and of square numbers 1, 1 + 3, 1 + 3 + 5, …, by the successive addition of numbers in arithmetical progression, called gnomons, is of geometric origin and goes back to Pythagoras¹ (570–501 B.C.):

      If the gnomons added are 4, 7, 10, … (of common difference 3), the resulting numbers 1, 5, 12, 22, … are pentagonal. If the common difference of the gnomons is m — 2, we obtain m-gonal numbers or polygonal numbers with m sides.

      In the cattle problem of Archimedes (third century B.C.), the sum of two of the eight unknowns is to be a triangular number (see Ch. XII).

      Speusippus,² nephew of Plato, mentioned polygonal and pyramidal numbers: 1 is point, 2 is line, 3 triangle, 4 pyramid, and each of these numbers is the first of its kind; also, 1 + 2 + 3 + 4 = 10.

      About 175 B.C., Hypsicles gave a definition of polygonal numbers which was quoted by Diophantus⁸ in his Polygonal Numbers, If there are as many numbers as we please beginning with one and increasing by the same common difference, then when the common difference is 1, the sum of all the terms is a triangular number; when 2, a square; when 3, a pentagonal number. And the number of the angles is called after the number exceeding the common difference by 2, and the side after the number of terms including 1. Given therefore an arithmetical progression with the first term 1 and common difference m — 2, the sum of γ terms is the γth m.

      The arithmetic of Theon of Smyrna⁴ (about 100 or 130 A.D.) contains 32 chapters. In Ch. 15, p. 41, the squares are obtained from 1 + 3 = 4, 1 + 3 + 5 = 9, etc. In Ch. 19, p. 47, the triangular numbers are defined to be 1, 1 + 2, 1 + 2 + 3, … In Ch. 20, p. 52, the squares are obtained as before and the pentagonal numbers are obtained by addition of 1, 4, 7, 10, … .

      Nicomachus⁵ (about 100 A.D.) gave the same definitions and results as did Theon of Smyrna and perhaps gave them slightly earlier. Ch. 12 gives the theorem on consecutive triangular numbers:

      also the corresponding theorem that the sum of the γth square and (γ — l)th triangular number is the γth pentagonal number, just as a pentagon is obtained by annexing a triangle to a square. He gave the generalization (apart from the notation):

      These theorems are illustrated by means of the following table:

      Each polygon equals the sum of the polygon immediately above it in the table and the triangle with 1 less in its side [triangle in the preceding column]; for example, heptagon 148 is the sum of hexagon 120 and triangle 28.

      Each vertical column is an arithmetical progression whose common difference is the triangle in the preceding column.

      In Ch. 13 he remarked that just as polygonal numbers arise by summing the simple arithmetical progressions, so by summing the polygonal numbers one obtains the like named pyramidal numbers, — triangular pyramid from the triangular numbers, pyramid with square base from the squares, etc., the base being the largest polygon.

      Plutarch,⁶ a contemporary of Nicomachus, gave the theorem that if we multiply a triangular number by 8 and add 1, we obtain a square:

      This theorem was given by Iamblichus⁷ (about 283–330 A.D.), who treated at length (pp. 82–176) polygonal and pyramidal numbers.

      Diophantus ⁸ (about 250 A.D.) generalized this theorem and proved by a cumbersome geometric method that

      and spoke of this result as a new definition of p equivalent to that of Hypsicles. Diophantus gave a rule for finding γ, equivalent to the solution of (1) for γ, and a rule for finding p equivalent to

      but did not give the equal simpler expression

      In fact, starting with (2), he gave a long geometric discussion to find the number of ways a given number can be polygonal, but made little headway before the abrupt termination of the fragment. G. Wertheim⁹ gave a lengthy continuation in the same geometric style which eventually leads to the geometric equivalent to (3) and remarked that we can readily find from (3) the ways in which a given number p can be polygonal: Express 2p as a product of two factors > 1 in all possible ways; call the smaller factor γ; subtract 2 from the larger factor and find whether or not the difference is divisible by γ – 1; if it is, the quotient is m – 2, and p Since m – 2 equals 2(p – γ)/[γ(γ 1, so that

      For example, if p = 36, then γ 8. Since γ divides 2p = 72, we have γ = 2, 4, 8, 3, 6, of which γ = 4 is excluded. We get

      In the Roman Codex Arcerianus¹⁰ (450 A.D.?) occur a number of special cases of the remarkable formula for pyramidal numbers

      , where the plus signs should be minus. M.Cantor¹¹ suggested the following probable derivation. By factoring the numerator of (2), we obtain

      As known by Archimedes (b. Syracuse about 287 B.C.),

      Hence

      The Hindu Aryabhatta¹² (b. 476 A.D.) gave the formula

      for the number of spheres in a triangular pile, and hence for the γof order 3, called also a tetrahedral number. The Hindus of his time knewwhence

      The above general formulas relating to polygonal and pyramidal numbers were collected about 983 A.D. by Gerbert¹⁴ (Pope Sylvester II).

      Yang Hui¹⁵ gave in his Suan-fa, 1261, the formulas

      for the sums of triangular numbers and squares.

      Chu Shih-chieh,¹⁶ in 1303, tabulated in the form of a triangle the binomial coefficients as far as those for eighth powers. This arithmetical triangle was known¹⁷ to the Arabs at the end of the eleventh century. Such a triangle was published by Petrus Apianus.¹⁸

      Many of the early arithmetics mentioned (some with fuller titles) in Vol. I, Ch. I, of this History, gave definitions and simple properties of

      polygonal numbers; for example, Boethius,¹⁹ G. Valla,²⁰ Martinus,²¹ Cardan,²² J. de Muris,²²a Willichius,²³ Michael Stifel,²⁴ who gave a table of figurate numbers (binomial coefficients), Faber Stapulensis,²⁵ and F. Maurolycus,²⁶ who gave

      and treated (pp. 32–74) polygonal numbers of the second order or central polygonal numbers (the pentagonal being 1, 6, 16, 31, 51, 76, …, when in the second are counted the vertices and center of a pentagon), as well as central pyramidal numbers (the pentagonal being 1, 7, 23, 54, 105, …). Also I. Unicornus,²⁷ and G. Henischiib.²⁸

      Johann Faulhaber²⁹ treated polygonal and pyramidal numbers.

      Johann Benzius³⁰ devoted twenty chapters to these and figurate numbers.

      J. Rudolff von Graffenried³¹ noted that

      the final number being 666 for γ = 6.

      C.G. Bachet³² wrote a supplement of two books to the Polygonal Numbers of Diophantus. The most important ones of his theorems (when expressed as formulas) are as follows:

      His II, 27, relates to the formula of Nicomachus⁵ (Ch. 20)

      from which follows the above formula II, 25, by addition (as in the Codex Arcerianus¹⁰). Fermat³³ generalized this proposition by introducing colonne: In the arithmetical progression 1, 1 + (m – 2), 1 + 2, (m – 2), … leading to m-gonal numbers, the first term 1 gives the first colonne; the sum of the next two terms diminished by m – 4 times the first triangular number 1 gives the second colonne 2m; the sum of the fourth, fifth and sixth terms diminished by m – 4 times the second triangular number 3 gives the third colonne 9m – 9; similarly, the fourth colonne is 8(3m – 4) and the rth is r² + r2(r – 1)(m – 2)/2. It follows (as noted by Editor Tannery) that the rth colonne is the product of the rth m-gonal number by r, and for m = 4 is r³. The term colonne was not coined by Fermat, as Tannery thought, but³⁴ was used by Maurolycus.²⁶

      J. Remmelin³⁵ noted that 666 (cf. Faulhaber²⁹) is a triangular number with the root 36, which is a square with the root 6, while 6 is a pronic number [of the form n² + n] whose base 2 is also a pronic number.

      Later we shall quote Bachet’s empirical theorem that any integer is the sum of four squares, made à propos of Diophantus IV, 31. In this connection Fermat³⁶ made the famous comment: I was the first to discover the very beautiful and entirely general theorem that every number is either triangular or the sum of 2 or 3 triangular numbers; every number is either a square or the sum of 2, 3 or 4 squares; either pentagonal or the sum of 2, 3, 4 or 5 pentagonal numbers; and so on ad infinitum, whether it is a question of hexagonal, heptagonal or any polygonal numbers. I can not give the proof here, which depends upon numerous and abstruse mysteries of numbers; for I intend to devote an entire book to this subject and to effect in this part of arithmetic astonishing advances over the previously known limits. But such a book was not published. Fermat³⁷ stated the theorem in a letter to Mersenne, Sept., 1636 (to be proposed to St. Croix); to³⁸ Pascal, Sept. 25, 1654, and Digby, June 19, 1658. The theorem was attributed to St. Croix by Descartes³⁹ in a letter to Mersenne, July 27, 1638. Descartes⁴⁰ gave an algebraic proof of Plutarch’s⁶ theorem that 8Δr + 1 = (2r + 1)². We shall often write Δr or Δ(r) for the rth triangular number r(r + 1)/2, Δ or Δ′ for any triangular number, □ for a sum of two, three or four squares.

      The rth figurate number of order n is the binomial coefficient

      is the ris the rIn a comment on the Polygonal Numbers of Diophantus, Fermat⁴¹ stated a theorem which, in the present notation, is

      the rth triangulo-triangular number.

      In April, 1638, St. Croix proposed to Descartes the problem: "Trouver un trigone [triangular number] qui, plus un trigone tétragone, fasse un tétragone [square], et de rechef, et que de la somme des côtés des tétragones résulte le premier des trigones et de la multiplication d’elle par son milieu le second. J’ai donné 15 et 120. J’attends que quelqu’un y satisfasse par d’autres nombres ou qu’il montre que la chose est impossible." The problem, without the example, was proposed to Fermat (Oeuvres, II, 63) in 1636, who did not solve it.

      Descartes⁴² understood a trigone tétragone to be the square Δ² of a triangular number, and proved that 15, 120 is the only solution if the problem is understood to require two triangular numbers such that, if either be added to the same Δ², the sum is a square; while if one is permitted to add both Δ² and a new Δ′² to the second required triangular number, the two triangular numbers may be taken to be 45 and 1035, since

      St. Croix did not admit the validity of Descartes’ solution, and probably meant a trigone tétragone to be a number both triangular and square (like 1, 36). The question would then be to find two numbers of the form n(n + 1)/2 such that, if a number both triangular and square be added to each, there result two squares; further, the sum of the square roots of these squares must equal the first required triangular number and must also be the first factor n used in forming the second triangular number. If, as seems intended, the numbers to be added to the triangular numbers are to be identical, the only solution is 15, 120. Cf. Gérardin.²²⁰

      Fermat⁴³ proposed to Frenicle the problem to find a number which shall be polygonal in a given number of ways. Neither gave a solution. [Cf. Euler,⁵⁹ end.]

      John Wallis(p. 143), the sum (called trianguli-pyramidal number) of the latter for r = 1, 2, …, l (p. 145), and the sum (called pyramidi-pyramidal number) of these last for l = 1, 2, … The values found are the expanded forms of

      so that his work amounts to a verification of cases of

      Frans van Schooten⁴⁵ quoted three of Bachet’s rules, proving one.

      On certain hexagons whose sum is a cube, see Frenicle⁶ of Ch. XXI.

      Fermat⁴⁶ proposed that Brouncker and Wallis find a proof of the proposition (which he himself could prove): There is no triangular number, other than unity, which is a biquadrate.

      Diophantus, IV, 44, desired three numbers which if multiplied in turn by their sum give a triangular number, a square, and a cube. Let the sum be x². Then the numbers are α(α + 1)/(2x², β²/x², γ²/x². Thus

      Take β = x² – 1. Then Δα = 2x² – γ³ – 1. But

      if x = (8γ³ + δ² + 7)/(8δ). Take γ = 2, δ = 1; then x = 9 and the desired numbers are 153/81, 6400/81, 8/81.

      Bachet convinced himself by trial that δ must be unity in order that α = (8γ³ + 7 – δ² – 2δ)/(4δ) be integral.

      Fermat remarked that "Bachet’s conclusion is not rigorous. Indeed, let γ be any number of the form 3n + 1, say γ = 7. To make

      and hence 16x² – 8.7³ – 7 = □, we may take the latter to be the square of 4x – 3 [whence x = 115, δ = 3]. Nothing prevents us from generalizing the method, taking instead of 3 any odd number and making a suitable choice of γ."

      G.Loria⁴⁷ remarked that the solution becomes evident if we replace x² by x; the problem did not require that the sum of the numbers be a square.

      Bachet³² (p. 274) proposed the problem to find five numbers which if multiplied in turn by their sum give a triangular number Δ, a square, a cube, a pentagonal number, and a biquadrate. The sum of the latter shall be x⁴. Let the square be (x² – 1)², the cube 8, the pentagonal number 5, and the biquadrate 1. Then Δ = 2x² – 15. Thus

      say (4x – 1)². Hence x = 15.

      René F. de Sluse⁴⁸ (1622–1685) employed the triangular number q, the square b² and cube z³. Then q + b² + z³ = □ = (b + n)², whence

      Hence we may assign any values to q, z³, n and find b. Likewise for Bachet’s generalization, we may assign any values to all five products other than the square b², and find b.

      A. Gérardin⁴⁹ noted that the simplest solution of Diophantus’ problem is furnished by the three numbers (x² + 1)/2, θ², x, with α = x², β = xθ, γ = x,

      Set x = 2H + 1. Then θ² – 2H² = – 1, with the solutions (H, θ) = (1, 1), (5, 7), (29, 41), etc., giving the numbers 5, 1, 3; 61, 49, 11; 1741, 1681, 59.

      René F. de Sluse⁵⁰ gave the table [cf. Stifel²⁴]

      in which the numbers (like 1, 3, 3, 1) in a diagonal are binomial coefficients, those in the third column are triangular numbers, those in the fourth column are pyramidal numbers with triangular base, those in the fifth are triangular pyramids of the second order.

      B. Pascal⁵¹ gave the same table and noted that any number in it is the sum of the numbers in the preceding column and hence (p. 504) is the sum of the number above it and that immediately to its left. He noted (p. 533) that n(n + 1) … (n + k – 1) is divisible by k!, the quotient being a figurate number.

      G. W. Leibniz⁵² gave a table formed by the diagonals (as 1, 2, 1) of the above table.

      J. Ozanam⁵³ found pairs* of triangular numbers 15 and 21, 780 and 990, 1747515 and 2185095, whose sum and difference are triangular. Their sides are 5 and 6, 39 and 44, 1869 and 2090. Polygonal numbers are treated in the English translation by C. Hutton, London, 1803, pp. 40–47, p. 60.

      Pierre Rémond de Montmort⁵⁴ cited special cases of (1), due to Dio-phantus.

      F.C. Mayer⁵⁵ defined generalized figurate numbers

      which for n = 2, 3, 4 include the polygonal numbers and the pyramidal numbers of the first and second kind, the number of sides being a + 2.

      L. Euler⁵⁶ investigated polygonal numbers which are also squares. The problem is a special case of that to make a quadratic function a square. The triangular numbers equal to squares are those with sides 0, 1, 8, 49, 288, 1681, 9800, … and equal the squares of 0, 1, 6, 35, 204, 1189, 6930, … The xth polygonal number with l sides is {(l – 2)x² – (l – 4)x}/2. To make it a square, set 2(l – 2)p² + 1 = q². Then the product of the polygonal number by 4 is the square of 0, (l – 4)p, 2(l – 4)pq, … if

      Euler gave a law for the derivation of any solution x in terms of two solutions. It remains to make the expressions (4) integers. For l = 5, q is to be chosen from 1, 5, 49, … and hence p from 0, 2, 20, …. The first fraction (4) is here* (1 – q)/6 and is an integer for q = 49, whence x = – 8. But Euler had previously stated that, for l > 4, q was to be taken negative. The value q = – 5 gives x = 1 and the pentagonal number 1.

      Euler⁵⁷ proved Format’s theorems that no triangular number except unity is a cube (since x⁶ ± y⁶ is not a square), and no triangular number x(x + 1)/2 > 1 is a fourth power. According as x is even or odd, x/2 or (x + 1)/2 must equal a fourth power m⁴, if the Δ is to be a fourth power. Thus 2m⁴ ± 1 = n⁴. But he had just proved that 2n⁴ ± 2 = □ only when n = 1, whence m = 0 or 1, x = 0 or 1.

      Abbé Deidier⁵⁸ gave the simplest properties of polygonal numbers and derived central polygonal numbers as follows: adding unity to the products of the triangular numbers 0, 1, 3, 6, 10, … by 3, 4 or 5, we get central triangular, square or pentagonal numbers, respectively.

      We shall now quote from the correspondence⁵⁹ between Euler and Gold-bach remarks on polygonal numbers, reserving for later use the comments in which the interest is chiefly on sums of squares. June 25, 1730 (p. 31), Euler noted that (x² + x)/2 equals (6/7)⁴ for x = 32/49, but said this does not disprove Fermat’s assertion that no (integral) triangular number is a biquadrate. Aug. 10, 1730 (p. 36), Euler noted that if

      the square of (a – b) is a triangular number with the side (a+b–2)/4 [evident since ab = 1]. Chr. Goldbach stated April 12, 1742 (p. 122) that 4mn – m – nΔ. Euler remarked May 8, 1742 (p. 123) that 4mn – m – n is not a heptagonal number. June 7, 1742 (p. 126), Goldbach inferred that every number is of the form 2 Δ ± □, and incorrectly (Euler, p. 134) that every number is a sum of three triangular numbers. Euler, June 30, 1742 (p. 133) noted that every number is of the form y² + y – x² = 2Δy – x². April 6, 1748 (pp. 447–9, 468), Goldbach stated that every number can be expressed in each of the eight forms

      June 25, 1748 (pp. 458–460), Euler gave the identity

      Hence [Fermat’s³⁶ theorem] every n is a sum of three Δ’s implies

      Euler expressed his belief that every number of the form 4n + of three squares, whence n = □ + □′ + 2Δ. Replacing n by 2n, we see that every n = □ + □′ + Δ. Euler gave fourteen such formulas. June 9, 1750 (p. 521), Euler remarked that an algebraic discussion of the theorem that any number n is a sum of three triangular numbers is of no help, since the theorem is not true if n ). Dec. 16,1752 (p. 597), Euler noted as facts, of which he had no proof, that every prime 8n + 1 or 8n + 3 is of the form x² + 2y², whence if n □+Δ, 8n+prime, and if n 2Δ + Δ′, 8n + prime. Also (p. 630), if n □ + 2 Δ, 4n + prime.

      April 3, 1753 (pp. 608–9), Euler treated the problem [of Fermat⁴³] to find a number z which is polygonal in a given number of ways. Let n be the number of sides of the polygonal number, x its root. Then

      Thus 2z must be divisible by x, and 2z – 2 by x – 1. Hence we desire two numbers differing by 2 which have divisors differing by 1. For example, 450 and 448 have such divisors 3 and 2, 5 and 4, 9 and 8, 15 and 14. Thus 225 is a square, 8–gon, 24–gon, and 76–gon.

      Euler⁶⁰ noted that, if 4n + 1 is a sum of two squares, 8n + 2 is a sum of two odd squares (2x + 1)², (2y + 1)², whence n = Δx + Δy. S. Réalis⁶⁰anoted that conversely this expression for n implies

      (1 – xk = ∑(– 1)jxp, where p = (3j²±j)/2 is a pentagonal number, and theorems by Legendre,²³Vahlen,¹⁵⁰ and von Sterneck,¹⁶⁹ on the partitions of N, in which an exceptional rôle is played by the N’s which are pentagonal or triangular.

      G. W. Kraft⁶¹ and A.G. Kastner⁶² proved that

      since (2²m – 1)/3 is an integer N.

      M. Gallimard⁶³ obtained central polygons by multiplying each term of 0, 1, 3, 6, 10, 15, … by the number n of angles of any polygon whatever and adding unity to each product. Given a central polygon, he treated the problem to find the number of angles if the side be given, or vice versa.

      L. Euler⁶⁴ proved that a number not the sum of a square and a triangular number Δ is composite; one not Δ + 2 Δ′ is composite.

      Nicolas Engelhard⁶⁵ treated Plutarch’s⁶ questions on triangular numbers.

      Elie de Joncourt⁶⁶ gave a table of triangular numbers N(N + 1)/2, N up to 20000, and showed how the table may be used to test if a number less than a hundred million is a square or not, and to extract square roots approximately.

      L. Euler⁶⁷ noted that, if N – ab = Δp + Δq + Δr and p – q = a – b, then N = Δp+b + Δp—a + Δr. N. Fuss, I, (pp. 191–6) also gave an incomplete argument to show that N is a sum of three triangular numbers if every integer N is. Let N – p = Δa + Δb + Δc and

      [a restriction]; then N = Δa—n + Δb+n + Δ c.He gave a similar incomplete discussion of the problem to express N as a sum of m m-gonal numbers, given that every integer < N is such a sum. He noted (p. 201) that 9n + 5, 8; 49n + 5, 19, 26, 33, 40, 47; 81n + 47, 74; etc., are not sums of two triangular numbers; thus, 49n + 19 = Δa + Δb would imply (2a + 1)² + (2b + 1)² = 8(49n + 19) + 2, whereas the factor 7 of the latter is not a divisor of a sum of two squares. L. Euler (p. 214) noted that ΔxΔy= ΔZ is satisfied* if px(y + 1) = 2qz, qy(x + 1) = p(z + 1); the resulting values of z are equal if {(2q² – p²)x + 2q²}y = p²x + 2pq. L. Euler (pp. 264–5, about 1775) noted that

      J. A. Euler⁶⁸ (the son of L. Euler) stated that to express every number as a sum of terms of 1², 3², 6², 10², 15², …, at least 12 terms are required.

      To express every number as a sum of figurate numbers

      at least a + 2n – 2 terms are necessary. Cf. Beguelin,⁷² Pollock,¹¹⁷Maillet.¹⁸¹–²

      L. Euler⁶⁹ remarked that Fermat’s³⁶ theorem that every integer is a sum of m m-gonal numbers would follow if we could prove that every integer occurs among the exponents in the expansion of the mth power of 1 + x + xm + x³m–3 + …, whose exponents are the m-gonal numbers. Fermat’s theorem that every integer is a sum of three triangular numbers would follow if it were shown that in

      all integers occur as exponents of x in the series for R.

      Euler⁷⁰ found squares which are triangular or pentagonal. If Δz = x², then y² = 8x² + 1 for y = 2z + 1. If (3z² – z)/2 = x², y² = 24x² + 1 for y = 6z – 1. If (3q² – q)/2 Δp – 1)² = 3x² – 2 for x = 2p + 1. Special solutions of the three equations y² = ax² + b are found by his general method of treating the latter (Ch. XII, below).

      Euleretc., can be resolved into three triangular numbers. There are no rational solutions x, y, z of

      Nicolas Beguelin⁷² attempted to prove Fermat’s theorem that every integer is a sum of s polygonal numbers of s sides. For s = d + 2, the latter are 0 and 1, A = d + 2, B = 3d + 3, C = 6d + 4, D = 10d + 5,…, a series whose second order of differences are d. Let t be the number of terms > 0 needed to produce a given sum e. For 1 < e < A, evidently t A – 1. Fore = A + εε A – 1, t A; fore = 2A + εε d – 2, t d. Next, let B . For e = B + εε A – 1, t A; for e = B + A + εε A – 2, t A; for the doubtful case e = B + A + A – 1, we replace B by its equal 2A + d – 1 and have e = 4A + d – 2, t = d + 2; finally, for e = B + 2A + εε d – 4, t d – 1. After this expansion of the argument by Beguelin, we are ready to admit that if e is in one of the intervals 1 to A, A to B, B to C, it is a sum of d + 2 or fewer terms 1, A, B. He treated four more intervals with a rapidly increasing number of doubtful cases for which linear relations between the polygonal numbers were employed, and found in every case that t d + 2. But he finally admitted (p. 405) that this method does not lead to a proof of the general theorem of Fermat.

      On p. 411, Beguelin stated without proof the erroneous generalization [cf. J. A. Euler,⁷³ L. Euler⁷³] that any number is the sum of at most t = d + 2n – 2 terms of the series

      a series whose nth order of differences are constant and equal to d. For n = 2, we have the case of polygonal numbers just considered. For n = 3, we have the pyramidal numbers Prd+2 for γ = 1, 2, 3,…; for n = 4, their sums, etc. For n = 4, d = 1, the series is 1, 5, 15, 35, 70,… and the theorem gives t = 7, whereas 8 terms are evidently required to produce the sum 64 (since 4 terms must be unity), as expressly mentioned on p. 412. Thus Beguelin contradicts himself in his generalization of Fermat’s theorem to pyramidal and figurate numbers.

      L. Euler⁷³ probably overlooked the last remark, since he stated that the unproved generalization merits great attention. He extended Beguelin’s tentative process to any series 1, A, B, …. We must employ A + n – 2 summands 1, A to produce nA – 1. Thus if

      we need A + n – 2 summands 1, A to produce all numbers 1, 2, …, B. Then

      Denote by {x} the least integer > x, and by t1 the number of terms 1. A needed to produce 1,…, B. Hence

      Bringing in also the summand B, let b be the least positive integer such that B + b requires t1 + 1 summands 1, A, B. If C < B + b, we need only t1 C. But if C B + b, let

      C from 1, A, B, we need

      summands. Next, bring in the summand C and let c be the least positive integer such that C + c requires t2 + 1 summands from 1, A, B, CD3 we need

      summands, etc. In the case of an infinite series 1, A, B, …, the process furnishes a lower limit to the number t of summands. Euler showed that, for series whose nth order of differences are constant, Beguelin’s rule is often quite erroneous, but did not treat the series 1, n + d, … of polygonal and pyramidal numbers.

      N. Beguelin. For,

      Adding 1, we conclude that 8n + . Hence 2n + . Since Lagrange had given in 1770 an independent proof of this theorem of Bachet, Beguelin next attempted, but failed completely, to deduce from it the result that every integer is a sum of three triangular numbers Δ. On p. 338, he gave the equivalent formulas

      He concluded without adequate proof that every number is a sum of a Δ and two squares, and also is Δ + 2Δ′ + 2Δ″ , later proved by Legendre ¹⁹of Ch. VII. A fitting sample of the lack of insight of Beguelin is furnished by his final theorem* (p. 368): If any number 4m + 0], it is composite [but 17 = 9 + 4 + 4 is prime]. Curiously enough, he supposed he had verified the theorem for all numbers < 200; but his tables (pp. 363–4) imply that he assumed that a number can be expressed in a single way as a sum of squares. On this he based a new proof that every prime 4m + .

      L. Euler⁷⁶ noted the result (5).

      Euler⁷⁷ noted that ½ is not the sum of three fractional triangular numbers (x² + x)/2, since 7 is not the sum of three odd squares (2x + 1)². But every number N is the sum of four fractional pentagonal numbers (3x² – x)/2, since

      To prove the theorem that any number is the sum of three integral triangular numbers Δ, it would be sufficient to show that the coefficient of every term xk in the expansion of

      is not zero; similarly for squares, pentagons, etc. [Euler⁶⁹]. Let the polygonal numbers with π sides be 0, α = 1, β = π, γ = 3π – 3, …, and denote by [n] the coefficient of xn in (1 + xa + xb + …)π. Euler proved by logarithmic differentiation the recursion formula

      F. W. Marpurg⁷⁸ treated (pp. 185–257) polygonal numbers, giving special cases of formula (1) of Diophantus, pyramidal numbers and central polygonal numbers, viz., unity more than the number of the angles and division points on m-gons drawn about a common mid point. Also (p. 307) poly-hedral numbers, the rth hexahedral, octahedral, dodecahedral and icosahedral being

      Euler⁷⁹ proved that (x² + x)/2 is a square y² only when

      For n = 0, 1, 2, we get x = 0, 1, 8; y = 0, 1, 6. We have the recursion formulas

      Certain squares x² which exceed (y² + y)/2 by unity are given by

      For n = 0, x = 1, y = 0; for n = 1, x = 4,y = 5. The recursion formula is

      A second series of solutions is obtained by use of these formulas for negative n’s. Thus x+1 = 2, y+1 = – 3; x+2 = 11, y+2 = – 16. Since the triangular number Δ+m equals Δm+1, we replace y = – m by m – 1 and get the sets of positive solutions 2, 2; 11, 15; etc.

      To find triangular numbers whose triples are triangular, Euler proved that 3(x² + x) = y² + y has only the solutions

      for n = 0, ± 1, ± 2, …. Examples are x = 1, y = 2; x = 5, y = 9.

      It was proposed as a prize problem in the Ladies’ Diary for 1792 to find n (n > 1) such that l² + 2² + … + n² = □. The sum is n(n + l)(2n + l)/6. T. Leybourn⁸⁰ took 2n + 1 = z², whence (z⁴ – l)/24 is to be a square y². Thus z⁴ = 24y² + 1 □ = (xy – l)², say. Thus y = 2x/(x² – 24) > 0. It is stated that x = 5 or 6. Since x = 6 is excluded, n = 24. C. Brady took n = 6r². Then the condition is (6r² + l)(12r² + 1) = □. Thus (9r² + l)² – (3r²)² = □, so that 9r² + 1 and 3r² equal the hypotenuse and one leg of a right triangle. Thus the other leg is 9r² – 1, whence r = 2, n = 24.

      A. M. Legendre⁸¹ proved Fermat’s theorems that no triangular number x(x + l)/2, except unity, is a fourth power or cube. For, in the first problem, x or x + 1 is of the form 2m⁴, whence either 1 = n⁴ – 2m⁴, contrary to 1 + 2m⁴ eqi_notequal □, or 1 = 2m⁴ – n⁴, m⁸ – n⁴ = (m⁴ – l)², contrary to p⁴ – n⁴ eqi_notequal □ unless m = 1 = x. In the second problem, one of 1 + x, x is a cube and the other the double of a cube, whence n³ ± 1 = 2m³, which is impossible if n eqi_notequal 1.

      C. F. Gauss⁸² proved by means of the theory of ternary quadratic forms that every number n = 8M + 3 is a sum of three odd squares, so that, by (5), M is a sum of three triangular numbers. The number of ways M can be so decomposed depends in a definite manner on the prime factors of n and the number of classes of binary quadratic forms of determinant – n.

      G. S. Klügeletc.

      John Gough⁸⁴ attempted to prove Fermat’s theorem that every number is a sum of m m-gonal numbers. P. Barlow⁸⁵ noted that the first three propositions by Gough are correct, but are not used in his defective proof of Fermat’s theorem, while various points are not proved, as the Cor. 2 to Prop. 4: every number is a sum of a limited number of polygonal numbers. As to Gough’s reply (pp. 241–5), Barlow⁸⁶ stated that the defense is on grounds not proved. As to the revised version by Gough⁸⁷, Barlow noted (p. 44) that the argument is correct and trivial to within 12 fines of the end; the proof is valid for numbers eqi_ltequal 3m, but not for those > 3m.

      E. Barruel⁸⁸ noted that sums of 1, 2, 3,… give the triangular numbers 1, 3, 6, … whose sums give the pyramidal numbers 1, 4, 10, 20,…, etc. Forming these sums, we get the general triangular and pyramidal numbers n(n + l)/2, n(n + l)(n + 2)/6, etc. Application is made to prove the ordinary rule for deriving a binomial coefficient from the preceding coefficients.

      F. T. Poselger⁸⁹ gave (pp. 19–31) various properties of numbers from the writings of Theon of Smyrna, and (pp. 32–60) gave algebraic expres-sions for polygonal and figurate numbers, with a discussion of arithmetical series of general order.

      P. Barlow⁹⁰ noted that, if N is a sum of five pentagons (3u² – u)/2, and M a sum of six hexagons 2x² – x, then

      In general, if P is a polygonal number of α + 2 sides, Fermat’s³⁶ theorem is equivalent to

      He erred¹⁰⁰,¹⁰⁷ (p. 258) in saying that no triangular number > 1 is pentagonal.

      J. Struve⁹¹ discussed figurate numbers (binomial coefficients).

      J. D. Gergonne⁹² noted that the number of terms of a polynomial of degree m in n unknowns is (m + n)! ÷(m! n!). If the latter be designated (m, n), then (m, n) = (m – 1, n)+ (m, n – 1).

      A. Cauchy⁹³ gave the first proof of Fermat's theorem that every number is a sum of m m-gonal numbers. The proof shows that all but four of the m-gons may be taken to be 0 or 1. The auxiliary theorems on sums of four squares will be quoted in Ch. VIII. In the simplified proof by Legendre,⁹⁴ the case m = 3 is not presupposed, as was done by Cauchy. Moreover, Legendre proved (p. 22) in effect that every integer > 28(m – 2)³is a sum of four m-gonal numbers if m is odd; while, for m even, every integer > 7(m – 2)³ is a sum of five m-gonal numbers one of which is 0 or 1.

      Cauchy⁹⁵ denoted the xth polygonal number of order m + 2 by

      and proved that if A, B,…, F are integers, not divisible by the odd prime p, there exist integers X1,…, xn, such that

      where n = m if m is even, and n = 2m if m is odd. The case m = 2 shows that there exist integral solutions of [Lagrange,⁹ etc., of Ch. VIII]

      L. M. P. Coste⁹⁶ showed that the problem to make two integral functions of one variable equal to polygonal numbers of a given order can be reduced to the problem to make two functions equal to squares. Let

      Then to make f1 and f2 equal to numbers P(Z), take Z = αz + a and Z = βz + b in the respective cases. We obtain a quadratic equation for α and one for β each linear in z. Solving for α and β we require that the quantities under the radical signs be squares, viz., 8pf1 + q² = □, 8pf2 + q2 = □. Next, if f1 and f2 are of the form 2a²pz² + Az + A′, use Z = 2az + α. We can make two quadratic functions equal to P(Z) if a particular solution is known.

      Several solvers⁹⁷ readily found two pentagonal numbers Px = (3x² – x)/2 and Pv whose sum and difference are triangular by solving

      A. M. Legendre⁹⁸ concluded from the formula

      that every integer N is a sum of four triangular numbers in σ (2N + 1) ways, where σ(k denotes the sum of the divisors of k. He gave an identity which shows the number of ways N is a sum of eight triangular numbers. Cauchy⁹⁹ gave (6); it was attributed to Jacobi by Bouniakowsky (see Vol. I, Ch. X¹²,¹⁹ of this History). Cf. Plana.¹²³

      Several¹⁰⁰ found numbers > Then m = 2p – 1, n =(1 + R)/6, where R² = 48p² – 24p + 1. Thus 1 + R = 6Kp, whence p = (2 – k)/(4 – 3k²). Take k = b/a. Then p is integral if 4a² – 3bwe get (2a, b) = (2, 1), (26, 15), (362, 209), (5042, 2911),…, whence p = 1, 143, 27693,…, so that answers are 1, 40755, 1533776801,

      J. Whitley¹⁰¹ found pairs of pentagonal numbers p, q whose sum and difference are pentagonal. The conditions are

      Hence z² = x² + y² – 1, v² = x² – y² + 1. Let x² = n² + m², y=² = 2nm + 1. Then z = n + m, v = n – m. Take n = r² – s², m = 2rs, whence x = r² + s². There remains the condition

      , which lead to larger numbers than those found by trial, using (r, s) = (3, 2), (6, 1), (8, 5), (13, 2), (13, 8), (19, 14). [But the resulting numbers p = 7, 37, 330,… are not pentagonal.] See Gill.¹⁰⁸

      C. G. J. Jacobi²² of Ch. VII gave in 1829 the result

      the exponents on the left being pentagonal numbers for m negative, and those on the right triangular. Polygonal numbers appear incidentally in Jacobi’s paper of 1848 [see Ch. III].

      J. Huntington,¹⁰² given a pentagonal number P = r(3r – l)/2 of n digits, found another number p also of n digits such that if p is prefixed to P there results a pentagonal number. Let x be the root of the latter. Then shall 10np = x(3x – l)/2. Taking p = x – r, we get

      For example, let r = 1; then n = 1, x = 6, p = 5 and 51 is pentagonal.

      A. Cauchy¹⁰³ defined triangular and pyramidal numbers as binomial coefficients.

      J. Baines¹⁰⁴ found two squares x², y² whose sum and difference are hexagonal. Take 8(x² – y²) + 1 = {2(x + y)± 1}², whence x = 3y ± 1. Then 8(x ² + y ²) + 1 = 80y² ± 48y + 9 = (ny ± 3)² determines y.

      A. Bernerie¹⁰⁵ gave a table of triangular numbers.

      A. Casinelli¹⁰⁶ noted that every triangular number is of one of the three forms

      also a sum of four Δ’s, and hence a sum of any number of Δ’s. By adding

      also a sum of four ’′s, and hence a sum of any number of ’′s. By adding the first two or the second and third equations, we get

      Also, (3m + 3)² = (m + 1)² + 2(2m + 2)². Hence every square is a sum of three squares or a sum of two squares and two A‚s. Further, every ’ is a sum of a square and two equal ’‚s. Next,

      and similarly for three or more Δ’s. Also, Δm + Δn – m(n + 1) = Δn–m

      C. Gill¹⁰⁷ found numbers both m-gonal and n-gonal, and the generalization

      where a, a′, b, b′ are given integers with no common factor. Take

      so that

      Let p′, q′ give a particular solution of the last equation such that

      are integers. Take p = p′t + abq′u, q = q′t + P′u. Then

      where F = t² – abu², and x = q(Bt + Abu)/F, y = q(At + Bau)/F. From the initial solution tΔ = 1, uΔ = 0, of F = 1, we get as usual the solution

      It remains only to find a solution p′, q′ of p² – abq² = – N, it suffices for our initial problem to note the solution p′ = a – a′, q′ = 1, for the case a – a′ = b – b′ then N = ab′ + ba′ – a′b′, A = B = 1. First, if m and n are both odd, we may take

      which have no common factor. Then a = b – b′

      But if m and n whence a – a′ = b – b′ = 1, and Pi" = Ti, satisfies the same recursion formula. Also,

      where e = 8 in the former case and e = 16 in the present case. For example, 1, 210, 40755 are both triangular and pentagonal, whereas Barlow⁹⁰ stated that this is true only for unity.

      Gill¹⁰⁸ found n-gonal numbers whose sum and difference are n-gonal, i. e., Px + Py = Pz, Px – Py = Pv where Px = (n – 2)x² – (n – 4)x. As a generalization, take Px = mx² – m′x, where m and m′ are relatively prime. The first condition is satisfied if

      Each of these linear equations is solved separately and the resulting x′s equated. The second of our conditions is treated similarly and the two sets of values of x and y are compared. But the resulting solution does not lead to convenient numbers. Another method is to assume that x = aw – h, y = bw, z = cw – h, where

      whence our first condition is satisfied. Thus take

      The second of our initial conditions now becomes

      where d = a – m′². Take 2mv – m′ = 2wt/u + 2mh + m′. We get w and then v rationally. By choice of the denominator t² – (d² – 2m′⁴) m²V² we get integral answers unless m′ = 0.

      Many¹⁰⁹ found two squares x², y² such that x² ± y² are pentagonal. Let 24(x² – y²) + 1 = {4(x + y) ± 1}², whence x = 5y 1. Then

      determines y. Again, to find pentagonal numbers p, q whose sum and difference are squares x2, y2 take 12(x² – y²) + 1 = {3(x + y) ± 1}² and 12(x² + y²) + 1 = (7 – yr/s)² whence a; = 7y ± 2.

      O. Terquem¹¹⁰ proved that no triangular number › 1 is a biquadrate.

      The ordinary definitions of polygonal and figurate numbers as sums of series were repeated by F. Stegmann,¹¹¹ George Peacock,¹¹² A. Transon,¹¹³H. F. Th. Ludwig¹¹⁴ Albert Dilling,¹¹⁵ and V. A. Lebesgue.¹¹⁶

      F. Pollock¹¹⁷ stated that every integer is a sum of at most 10 odd squares, and a sum of at most 11 triangular numbers 1,10, 28, 55,… of rank 3n+l, while 5, 7, 9, 13, 21, 11 terms are needed to express every number as a sum of tetrahedral, octahedral, cubic, icosahedral, dodecahedral, and squares of triangular, numbers. Legendre had proved that 8n + 3 is a sum of three odd squares, each being 8 ’ + 1. Pollock gave the generalization that, if Fx is any figurate number of order x, 8FX + 3 is a sum of 3 or 3 + 8,…, or 3 + 8n terms of a series whose general term is 8Fy + 1.

      V. Bouniakowsky¹¹⁸ employed (1) and (9) of Vol. 1, Ch. X, of this History, to prove that every odd pentagonal number can be expressed as a sum of another pentagonal number and either a square or the double of a square; every odd square not a triangular number is a sum of double a triangular number and either a square or the double of a square. Similarly,

      the factor (1, 2) denoting 1 or 2.

      F. Pollock¹¹⁹ stated without proof that any integer between two consecutive triangular numbers is the sum of four triangular numbers the sum of whose bases is constant.

      J. B. Sturm¹²⁰ gave the relations

      V. A. Lebesgue¹²¹ gave two proofs of the final theorem under Wallis.⁴⁴

      J. Liouville¹²² proved readily that the only forms aΔ + bΔ′ + cΔ″ which represent all numbers, where the Δ′s are triangular numbers and a, b, c are positive integers, are Δ + Δ′ + cΔ″ (c = 1, 2, 4, 5) and Δ + 2Δ′ + dΔ″ (d . The case c = 1 was treated by Gauss.⁸² Next,

      proves the case c = 2. Next,

      or case d = 2. Next, as shown by Gauss,

      The proofs for the remaining cases c = 5 and d = 3 are longer.

      J. Plana¹²³ wrote ξ⁴ for the left member of (6). By expanding the second member as a power series in q and examining the earlier terms, he verified that

      where σ(k) is the sum of the divisors of k. Hence any integer n is a sum of 4 triangular numbers in σ(2n + 1) ways. Give to ξ³ the notation of a power series in q, multiply it by £ aod compare with the above series for ξ⁴; we get a recursion formula for the coefficients of ξ³. He states without proof that the coefficient of every power of q is not zero, and so concludes that every integer is a sum of three triangular numbers.

      F. Pollock¹²⁴ verified for small values that any number may be expressed in the form s – s′, where s and s′ are sums of two triangular numbers. Now s is always the sum of a square and the double of a triangular number. Thus the theorem is that

      represents any number. Take p² – c² – c + q as the number. Then

      Double and add unity. Thus A = M + 2q, where

      Since q is arbitrary, it is concluded that any odd number can be represented by either of the forms A or M. But M is the sum of four squares.

      by (7). As before,

      represents any odd number 2n + 1. But a² + a + b² is the sum of two triangular numbers. Hence n is the sum of three triangular numbers.

      Pollock⁵³ of Ch. VIII noted that the theorem that every number 4n + 2 is a sum of four squares implies that every integer n is a sum of four ’′s.

      J. Liouville¹²⁵ considered the partition of any number into a sum of ten triangular numbers.

      S. Bills¹²⁶ solved ’x + ’y = ’a by setting y = a – xr/s and finding x rationally.

      E. Lionnet stated and V. A. Lebesgue and S. R£alis¹²⁷ proved that every integer is a sum of a square and two ’′s, also a sum of two squares and a ’.

      A. Hochheim¹²⁸ gave linear relations between polygonal and polyhedral numbers.

      S. R6alis¹²⁹ proved that every integer is a sum of four numbers of the form (3z² ± z)/2 and also of four of the form 2z² ± z, i. e., pentagons and hexagons extended to negative arguments. Use was made of the theorems that any odd number co is a sum of four squares the algebraic sum of whose roots is 1 or 3, and its double 2o> is a sum of four squares the algebraic sum of whose roots is zero. Further, every odd number divisible by h, or the double of every odd number divisible by the even number h, is a sum of four polygonal numbers of order h + 2, extended to negative arguments.

      E. Lucas¹³⁰ stated that [cf. Leybourn⁸⁰] 1² +… + n² is a square only when n = 24 [and n = 1], and is never a cube or fifth power. A triangular number [> 1] is never a cube, biquadrate or fifth power [Euler⁵⁷]. No pyramidal number is a cube or fifth power, or a square with the exception of

      Hence except for these and for the pile 24–25–49/6 = 70² with a square base, no pile of bullets with a triangular or square base contains a number of bullets equal to a square, cube or fifth power.

      Lucas¹³¹ stated and proved incompletely that the [pyramidal] number x(x+1)(2x+1)/6 of bullets in a pile, whose base is a square with x to a side, is a square only when x = 1 or 24 (see papers 130, 132, 137–8).

      T. Pepin¹³² noted that one case of Lucas’ proof of the last result leads to an equation 9γ⁴ – 12f²r² – 4f⁴ = R², not treated by Lucas when f and R are divisible by 3. Pepin found an infinitude of solutions in this case. G. N. Watson¹³²a noted the solution r = 5, f = 3, R = 51, and¹³²b proved Lucas’¹³¹ theorem by use of elliptic functions.

      Lucas¹³³ stated that the number of bullets in a pile with a square or triangular base is never a cube or fifth power. Moret-Blanc¹³⁴ gave a proof.

      Moret-Blanc¹³⁵ noted that the tetrahedral number n(n + 1)(n + 2)/6 is a square for n = 1, 2, 48. Lucas stated that it is a square only then, a fact proved by A. Meyl.¹³⁶

      E. Fauquembergue¹³⁷ and N. Alliston¹³⁸ proved that 1² + … + n□ if n > 24. Cf. Lucas¹³¹ and the papers cited on p. 26.

      For analogous theorems on sums of consecutive squares or the sum of the squares of the first n odd numbers see papers 70, 76, 81, 86, 87, 100, and 103 of Ch. IX, and Brocard⁹² of Ch. XXIII.

      W. Göring¹³⁹ proved by use of infinite series that 2Δ + 6Δ’ + 1 can always be represented by the form a² + 3b².

      J. W. L. Glaisher¹⁴⁰ noted that every representation of an odd number as a sum of an even square and two triangular numbers corresponds to a representation in which the square is odd, since

      for p,q both even or both odd, with a similar identity if one is even and the other odd.

      Glaisher¹⁴¹ stated that every triangular number is a sum of three pentagonal numbers.

      D. Marchand¹⁴² noted the relations

      Marchand¹⁴³ gave identities like

      where y = x² + x, and (p. 105) discussed triangular numbers which are squares.

      E. Lucas/(Δ1 + … + Δn is a square.

      S. Réalis, E. Catalan and others¹⁴⁵ investigated numbers simultaneously squares and triangular. S. Réalis stated and E. Cesàro¹⁴⁶ proved that the square of every odd multiple of 3 is a difference of two Δ’s prime to 3, 9(2n + 1)² = Δ(9n + 4) – Δ(3n + 1). D. Marchand¹⁴⁷ gave the generalization that the square of any odd number is the difference of two relatively prime triangular numbers (with sides 3x + 1 and x). C. Henry¹⁴⁸proved a like result for the product of any odd square by any number.

      S. Réalis¹⁴⁹ stated that the theorem that every integer n is a sum of three Δ ’s implies that n is a sum of four Δ ’s of which two are consecutive and that n is a sum of four Δ ’s two of which are equal.

      E. Lucas¹⁵⁰ listed values of A for which xy(x + y) = Az³ has no distinct rational solutions ≠ 0. Taking y = 1 and y = x + 1, we obtain theorems on triangular numbers and numbers x(x + l)(2x + 1).

      Lucas stated and Moret-Blanc¹⁵¹ proved that l² + … + x² = ky² are impossible if k = 2, 3, 6.

      S. Roberts¹⁵² proved by use of y² – 2z² = 1 [Euler⁷⁰] that the Δ’s which are squares are

      J. Neuberg stated and E. Cesàro¹⁵³ proved that the sum of the squares of n + 1 consecutive integers, beginning with the 2nth triangular number, equals the sum of the squares of the n succeeding integers, each being divisible by 1² + … + n². Cf. Dostor⁷⁵ of Ch. IX.

      E. Lionnet¹⁵⁴ stated that unity is the only triangular number Δ which equals the sum of the squares of two consecutive integers; 10 is the only Δ equal to the sum of the squares of two consecutive odd integers; when Δ is a product of two consecutive integers of which the least is double a triangular number, then 4 Δ + 1 (and its square root) is a sum of squares of two consecutive integers.

      Moret-Blanc¹⁵⁵ proved the preceding theorems stated by Lionnet.

      E. Cesàro¹⁵⁶ noted that no triangular number ends with 2, 4, 7, 9.

      S. Réalis¹⁵⁷ noted that

      E. Lionnet¹⁵⁸ noted that 0, 1, 6 are the only Δ’s whose squares are Δ’s. He stated and E. Cesaro¹⁵⁹ proved that there is at least one and at most two Δ’s between any two consecutive squares ≠ 0; at most one square between two consecutive Δ’s; if there are exactly two Δ’ between (a + 1)² and (a + 2)², where a< 0, there is just one Δ between a² and (a + 1)², and just one Δ between (a + 2)² and (a + 3)².

      E. Cesàro¹⁶⁰ denoted by Δ(n) the number of the first 2n triangular numbers which are relatively prime to n. Let Ψ n) be the number of products 1.2, 2.3, 3.4, …, n(n + 1) which are prime to n. Then if v is the largest odd divisor of n,

      The mean value of Δ(n) is three times that of Ψ(n). He found that the probability that two triangular numbers taken at random shall be relatively prime is

      Cesàro¹⁶¹ stated and E. Fauquembergue¹⁶¹ proved that 5 and 17 are the only integers whose cubes diminished by 13 are quadruples of triangular numbers.

      G. de Rocquigny¹⁶² noted that, if k = a² + b² + a + b)/2,

      S. Réalis¹⁶³ used the known fact that, if p is a product of primes 8q + 1, 2x² + y² = p has integral solutions. Thus

      so that 3q = 2Δ + Δ′.

      Réalis¹⁶⁴ gave various sums like

      E. Cesàro¹⁶⁵ noted that (n⁵ – l)/4 = Δp + Δq,n ≠ implies that 2p + 1 or 2q + 1 is composite.

      S. Tebay and otherswhich when increased by a² is

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