A258150 Triangle of Fibonacci's congruum (congruous) numbers divided by 24 based on primitive Pythagorean triangles. Areas divided by 6 of these triangles.

1, 0, 5, 10, 0, 14, 0, 35, 0, 30, 35, 0, 0, 0, 55, 0, 105, 0, 154, 0, 91, 84, 0, 220, 0, 260, 0, 140, 0, 231, 0, 390, 0, 0, 0, 204, 165, 0, 455, 0, 0, 0, 595, 0, 285, 0, 429, 0, 770, 0, 935, 0, 836, 0, 385, 286, 0, 0, 0, 1190, 0, 1330, 0, 0, 0, 506

Offset: 2

Author

Wolfdieter Lang, Jun 11 2015

Keywords

• nonn
• easy
• tabl

Comments

The problem is: given a square, find a positive integer that, whether added to or subtracted from that square, yields a square. That is, both x^2 + C = y^2 and x^2 - C = z^2. Equivalently: z^2 + C = x^2 and x^2 + C = y^2 (squares in arithmetic progression). This is treated in Fibonacci's 'The book of squares' (Liber Quadratorum (1225) but for rational x,y,z). See the Sigler reference, Proposition 14, pp. 53-74 (note that the formulation of this problem on p. 53 is not correct, 'from a square' should read 'from the same square'). See also van der Waerden, pp. 40-42, and A. Weil, pp. 13-14. The desired number C is called a congruum by Fibonacci (a congruous number in Sigler's translation).

For the history of this problem, see Dickson, pp. 459-472 (he uses the (misleading) term congruent number).

The following solution is based on primitive Pythagorean triangles. (Fibonacci's solution is based on sums of odd squares.) The triangle T(n, m) = 24*C(n, m) will have 0's for those (n, m) not leading to primitive Pythagorean triples.

Addition of the two equations, substitution of y = u + v > 0 and z = |u - v| and division by 2 leads to x^2 = u^2 + v^2. Consider primitive Pythagorean triples (u, v, x) with even v which are pairwise relatively prime. Then also GCD(u,v,x) = 1. A common factor f for u, v and x would lead to a multiplication by f^2 on both sides of the two equations. For primitive Pythagorean triples see A249866. One has u = n^2 - m^2, v = 2*n*m and x = n^2 + m^2 with GCD(n, m) = 1 , n > m >= 1, n + m odd. Then C = C(n, m) = 4*n*m*(n^2 - m^2) = 2*v(n, m)*u(n, m). This is four times the area of the Pythagorean triangle. C is divisible by 4! = 24 (see A020885). Define T(n, m) = C(n, m)/4!, for 2 <= m + 1 <= n. This is the area of the corresponding primitive Pythagorean triangle divided by 6.

The corresponding x = x(n, m), y = y(n, m) and z(n, m) number triangles are given in A222946, A225949 and A258149 respectively.

T(n, m) = n*m*(n^2 - m^2)/6, for m = 1, 2, ..., n-1, has for n >= 2 the minimum value at m = 1, and the next largest value appears for n >= 3 at m = n-1. Note all (n, m) pairs are considered here. The proof of the first part is easy. The proof of T(n, m) - T(n, n-1) > 0, for m = 2, 3, ..., n-2, and n >= 3, is equivalent to n^2*(m-2) + 3*n > m^3 +1 and this is easy to prove with n >= m+2 and m >= 2. Therefore the triangle T(n, m) with 0's attains for even n the smallest nonzero row entry at m = 1, and for odd n the smallest nonzero row entry appears at m = n-1 (last entry).

This allows us to find (after solution of two cubic equations for even and odd n, named ne = ne(N) and no = no(N)) a row number nmin(N) = max(ne(n), no(N)) such that N will not appear in any row n > nmin(N).

The original problem posed to Fibonacci by Giovanni di Palermo (Master John of Palermo) was to find a [rational] square that when increased or decreased by 5 gives a square. Fibonacci gave the solution in his Liber Quadratorum in Proposition 17 (see Sigler, pp. 77-81) as x^2 = (41/12)^2 = 1681/144, y^2 = (49/12)^2 = 2401/144 and z^2 = (31/12)^2 = 961/144. This corresponds to the integer quartet (C; x, y, z) = (720; 41, 49, 31) corresponding to the primitive Pythagorean triple [9, 40, 41]. See the examples for (n, m) = (5, 4).

The numbers without zeros, in nondecreasing order, are given in A020885 = A024406/6.

Comments from Eric Snyder, Feb 07 2023: (Start)

If m+n > 3 and not divisible by 3, then m+n | T(n,m).

Additionally, if 2n-1 > 3 and not divisible by 3, then 2n-1 = 6k+-1, and T(n,n-1) = (2n-1)*P(-+k), where P(-+k) is a generalized pentagonal number (A001318). For example, T(6,5) = 11*P(-2) = 11*5.

T(n,n-1) = A000330(n-1) for n>=2. (End)

Examples

			The triangle T(n, m) begins:
n\m   1   2   3   4    5   6    7   8   9  10    11
2:    1
3:    0   5
4:   10   0  14
5:    0  35   0  30
6:   35   0   0   0   55
7:    0 105   0 154    0  91
8:   84   0 220   0  260   0  140
9:    0 231   0 390    0   0    0 204
10: 165   0 455   0    0   0  595   0 285
11:   0 429   0 770    0 935    0 836   0 385
12: 286   0   0   0 1190   0 1330   0   0   0   506
...
The smallest nonzero number for each row with even n is T(n, 1), and for odd n it is T(n, n-1).
The above mentioned nmin(N) will for N = 300 be 12.
Therefore, no number > 300 will appear for rows with n > 12.
-----------------------------------------------------
The corresponding quartets (C; x, y, z) are:
n=2:  (24; 5, 7, 1),
n=3:  (120; 13, 17, 7),
n=4:  (240; 17, 23, 7), (336; 25, 31, 17),
n=5:  (840; 29, 41, 1), (720; 41, 49, 31),
n=6:  (840; 37, 47, 23), (1320; 61, 71, 49),
n=7:  (2520; 53, 73, 17), (3696; 65, 89, 23),
      (2184; 85, 97, 71),
n=8:  (2016; 65, 79, 47), (5280; 73, 103, 7),
      (6240; 89, 119, 41), (3360; 113, 127, 97),
n=9:  (5544; 85, 113, 41), (9360; 97, 137, 7),
      (4896; 145, 161, 127),
n=10: (3960; 101, 119, 79), (10920; 109, 151, 31),
      (14280; 149, 191, 89), (6840; 181, 199, 161),
n=11: (10296; 125, 161, 73), (18480; 137, 193, 17),
      (22440; 157, 217, 47), (20064; 185, 233, 119),
      (9240; 221, 241, 199),
n=12: (6864; 145, 167, 119), (28560; 169, 239, 1),
      (31920; 193, 263, 73), (12144; 265, 287, 241),
...
-----------------------------------------------------
The corresponding primitive Pythagorean triples
(u, v, x) are:
n=2:  [3, 4, 5],
n=3:  [5, 12, 13],
n=4:  [15, 8, 17], [7, 24, 25],
n=5:  [21, 20, 29],[9, 40, 41],
n=6:  [35, 12, 37], [11, 60, 61],
n=7:  [45, 28, 53], [33, 56, 65],
      [13, 84, 85],
n=8:  [63, 16, 65], [55, 48, 73],
      [39, 80, 89], [15, 112, 113],
n=9:  [77, 36, 85], [65, 72, 97],
      [17, 144, 145],
n=10: [99, 20, 101], [91, 60, 109],
      [51, 140, 149], [19, 180, 181],
n=11: [117, 44, 125], [105, 88, 137],
      [85, 132, 157], [57, 176, 185],
      [21, 220, 221],
n=12: [143, 24, 145], [119, 120, 169],
      [95, 168, 193], [23, 264, 265],
...
		

References

• L. E. Dickson, History of the Theory of Numbers. Carnegie Institute Public. 256, Washington, DC, Vol. 2, 1920, pp. 459-472.
• L. E. Sigler, Leonardo Pisano, Fibonacci, The book of squares, Academic Press, 1987.
• B. L. van der Waerden, A History of Algebra, Springer, 1985, pp. 40-42.
• André Weil, Number Theory, An approach through history, From Hammurapi to Legendre, Birkhäuser, 1984, pp. 13-14.

Crossrefs

Cf. A020885, A024365, A024406, A249866, A222946, A225949, A258149.

Programs

Mathematica

T[n_, m_] /; 2 <= m+1 <= n && OddQ[n+m] && CoprimeQ[n, m] := n*m*(n^2 - m^2)/6; T[, ] = 0; Table[T[n, m], {n, 2, 12}, {m, 1, n-1}] // Flatten (* Jean-François Alcover, Jun 16 2015, after given formula *)

Formula

T(n, m) = n*m*(n^2 - m^2)/6 if 2 <= m+1 <= n, n+m odd, GCD(n, m) = 1 and 0 otherwise.