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

%I A258150 #27 Feb 26 2023 20:42:15
%S A258150 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,
%T A258150 0,140,0,231,0,390,0,0,0,204,165,0,455,0,0,0,595,0,285,0,429,0,770,0,
%U A258150 935,0,836,0,385,286,0,0,0,1190,0,1330,0,0,0,506
%N A258150 Triangle of Fibonacci's congruum (congruous) numbers divided by 24 based on primitive Pythagorean triangles. Areas divided by 6 of these triangles.
%C A258150 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).
%C A258150 For the history of this problem, see Dickson, pp. 459-472 (he uses the (misleading) term congruent number).
%C A258150 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.
%C A258150 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.
%C A258150 The corresponding x = x(n, m), y = y(n, m) and z(n, m) number triangles are given in A222946, A225949 and A258149 respectively.
%C A258150 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).
%C A258150 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).
%C A258150 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).
%C A258150 The numbers without zeros, in nondecreasing order, are given in A020885 = A024406/6.
%C A258150 Comments from _Eric Snyder_, Feb 07 2023: (Start)
%C A258150 If m+n > 3 and not divisible by 3, then m+n | T(n,m).
%C A258150 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.
%C A258150 T(n,n-1) = A000330(n-1) for n>=2. (End)
%D A258150 L. E. Dickson, History of the Theory of Numbers. Carnegie Institute Public. 256, Washington, DC, Vol. 2, 1920, pp. 459-472.
%D A258150 L. E. Sigler, Leonardo Pisano, Fibonacci, The book of squares, Academic Press, 1987.
%D A258150 B. L. van der Waerden, A History of Algebra, Springer, 1985, pp. 40-42.
%D A258150 André Weil, Number Theory, An approach through history, From Hammurapi to Legendre, Birkhäuser, 1984, pp. 13-14.
%F A258150 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.
%e A258150 The triangle T(n, m) begins:
%e A258150 n\m   1   2   3   4    5   6    7   8   9  10    11
%e A258150 2:    1
%e A258150 3:    0   5
%e A258150 4:   10   0  14
%e A258150 5:    0  35   0  30
%e A258150 6:   35   0   0   0   55
%e A258150 7:    0 105   0 154    0  91
%e A258150 8:   84   0 220   0  260   0  140
%e A258150 9:    0 231   0 390    0   0    0 204
%e A258150 10: 165   0 455   0    0   0  595   0 285
%e A258150 11:   0 429   0 770    0 935    0 836   0 385
%e A258150 12: 286   0   0   0 1190   0 1330   0   0   0   506
%e A258150 ...
%e A258150 The smallest nonzero number for each row with even n is T(n, 1), and for odd n it is T(n, n-1).
%e A258150 The above mentioned nmin(N) will for N = 300 be 12.
%e A258150 Therefore, no number > 300 will appear for rows with n > 12.
%e A258150 -----------------------------------------------------
%e A258150 The corresponding quartets (C; x, y, z) are:
%e A258150 n=2:  (24; 5, 7, 1),
%e A258150 n=3:  (120; 13, 17, 7),
%e A258150 n=4:  (240; 17, 23, 7), (336; 25, 31, 17),
%e A258150 n=5:  (840; 29, 41, 1), (720; 41, 49, 31),
%e A258150 n=6:  (840; 37, 47, 23), (1320; 61, 71, 49),
%e A258150 n=7:  (2520; 53, 73, 17), (3696; 65, 89, 23),
%e A258150       (2184; 85, 97, 71),
%e A258150 n=8:  (2016; 65, 79, 47), (5280; 73, 103, 7),
%e A258150       (6240; 89, 119, 41), (3360; 113, 127, 97),
%e A258150 n=9:  (5544; 85, 113, 41), (9360; 97, 137, 7),
%e A258150       (4896; 145, 161, 127),
%e A258150 n=10: (3960; 101, 119, 79), (10920; 109, 151, 31),
%e A258150       (14280; 149, 191, 89), (6840; 181, 199, 161),
%e A258150 n=11: (10296; 125, 161, 73), (18480; 137, 193, 17),
%e A258150       (22440; 157, 217, 47), (20064; 185, 233, 119),
%e A258150       (9240; 221, 241, 199),
%e A258150 n=12: (6864; 145, 167, 119), (28560; 169, 239, 1),
%e A258150       (31920; 193, 263, 73), (12144; 265, 287, 241),
%e A258150 ...
%e A258150 -----------------------------------------------------
%e A258150 The corresponding primitive Pythagorean triples
%e A258150 (u, v, x) are:
%e A258150 n=2:  [3, 4, 5],
%e A258150 n=3:  [5, 12, 13],
%e A258150 n=4:  [15, 8, 17], [7, 24, 25],
%e A258150 n=5:  [21, 20, 29],[9, 40, 41],
%e A258150 n=6:  [35, 12, 37], [11, 60, 61],
%e A258150 n=7:  [45, 28, 53], [33, 56, 65],
%e A258150       [13, 84, 85],
%e A258150 n=8:  [63, 16, 65], [55, 48, 73],
%e A258150       [39, 80, 89], [15, 112, 113],
%e A258150 n=9:  [77, 36, 85], [65, 72, 97],
%e A258150       [17, 144, 145],
%e A258150 n=10: [99, 20, 101], [91, 60, 109],
%e A258150       [51, 140, 149], [19, 180, 181],
%e A258150 n=11: [117, 44, 125], [105, 88, 137],
%e A258150       [85, 132, 157], [57, 176, 185],
%e A258150       [21, 220, 221],
%e A258150 n=12: [143, 24, 145], [119, 120, 169],
%e A258150       [95, 168, 193], [23, 264, 265],
%e A258150 ...
%t A258150 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 *)
%Y A258150 Cf. A020885, A024365, A024406, A249866, A222946, A225949, A258149.
%K A258150 nonn,easy,tabl
%O A258150 2,3
%A A258150 _Wolfdieter Lang_, Jun 11 2015