Congruum ProblemDate: 04/04/2002 at 18:06:35 From: Dr. Allan Subject: Congruum problem In researching suitable problems for my students I have found a reference to Fibonacci and his congruum problem. Something has me stumped; I hope one or more of you can tell me what I am doing wrong: Quote from the article at the MacTutor History of Mathematics archive at St. Andrews University: http://www-history.mcs.st-and.ac.uk/history/Mathematicians/Fibonacci.html "[Fibonacci] defined the concept of a congruum, a number of the form ab(a + b)(a - b), if a + b is even, and 4 times this if a + b is odd where a and b are integers. Fibonacci proved that a congruum must be divisible by 24 and he also showed that for x,c such that x^2 + c and x^2 - c are both squares, then c is a congruum. He also proved that a square cannot be a congruum." With x=15 and c=216, we get the two squares 441 and 9, meaning that 216 should be a congruum. Thus we should be able to find numbers a and b such that 216 = ab(a+b)(a-b) if a+b is even or 54 = ab(a+b)(a-b) if a+b is odd Now, b must be smaller than a and d:= ab(a+b)(a-b) is increasing in both a and b. Let's look at the possible scenarios. Say a+b is odd. We want d to equal 54. Our triples (b,a,d) give us (1,2,6) (1,4,60); thus b=1 is not possible (2,3,30) (2,5,210); thus b=2 is not possible (3,4,84); thus b = 3 and b > 3 is not possible (since d is increasing); thus a+b odd is not possible Say a+b is even. We want d to equal 216. Our triples (b,a,d) give us (1,5,120) (1,7,336); thus b=1 is not possible (2,4,96) (2,6,3849; thus b=2 is not possible (3,5,360); thus b = 3 and b > 3 is not possible (since d is increasing); thus a+b even is not possible So, because 15^2 + 216 = 21^2 and 15^2 - 216 = 3^2 we should have that 216 is a congruum, but we cannot put it in the form ab(a+b)(a-b). What is wrong with this argument? Sincerely, Dr. Allan Date: 04/05/2002 at 16:04:44 From: Dr. Paul Subject: Congruum problem I went to the library this morning and picked up a copy of Ore's _Number Theory and its History_. The solution to the congruum problem is listed on pages 188-193 and I found it very interesting reading. What follows is essentially Ore's solution with my commentary added where I felt Ore's details were (probably intentionally) lacking sufficient explanation: We want to find a number x such that simultaneously: x^2 + h = a^2, x^2 - h = b^2 (equation 8-16) and determine for which h rational solutions x can exist. We shall first determine the solutions in integers, and this depends, as we shall see again, on the Pythagorean triangle. When the second equation in (8-16) above is subtracted from the first, one has: 2*h = (a^2 - b^2) = (a-b)*(a+b) (8-17) Since the left-hand side is even, a and b must both be odd or both be even. Therefore, a - b is even. Hence a - b = 2*k and k must be a divisor of h since according to (8-17) we have: 2*h = 2*k*(a+b) or h = k*(a+b) which implies k divides h. It follows that (a + b) = h/k and by adding and subtracting the last two equations, one finds a = h/(2*k) + k, b = h/(2*k) - k When these two expressions are substituted into the original equations (8-16), we obtain: x^2 + h = [h/(2*k) + k]^2 = [h/(2*k)]^2 + h + k^2 and x^2 - h = [h/(2*k) - k]^2 = [h/(2*k)]^2 - h + k^2 Adding these equations and removing a common factor of two gives: x^2 = [h/(2*k)]^2 + k^2 In our case of interest, we have: x = 15 h = 216 a = 21 b = 3 k = 9 Therefore, the three numbers x, h/(2*k), k form a Pythagorean triangle. Determine the lesser of h/(2*k) and k (in this case k = 9 < 12 = 216/18 = h/(2*k)) so that we can write: x = t*(m^2 + n^2) k = t*(m^2 - n^2) (**) h/(2*k) = 2*m*n*t where t is some integer and the expressions in m and n define a primitive solution of the triangle. In actuality, we have: x = 15 k = 9 h = 216 so clearly, t = 3 and this forces m = 2, n = 1 As an aside, if h/(2*k) > k then we pick h/(2*k) = t*(m^2 - n^2) k = 2*m*n*t Now we take the product of the last two expressions in (**) above to obtain as the general solution to (8-16) x = t*(m^2 + n^2), h = 4*m*n*(m^2 - n^2)*t^2 (8-18) Now we make a slight reduction to this solution. Let us suppose that we have a solution x of (8-16), where x has the factor t and h at the same time has the factor t^2 (this is the case in our example since x = 15 is divisible by three and h = 216 is divisible by 9). Then we can write: x = x_1 * t, h = h_1 * t^2 From (8-16), we obtain: [(x_1)^2 * t^2] + (h_1 * t^2) = a^2 and [(x_1)^2 * t^2] - (h_1 * t^2) = b^2 Then a^2 and b^2 are both divisible by t^2 so a and b are both divisible by t: a = a_1 * t, b = b_1 * t This gives: [(x_1)^2 * t^2] + (h_1 * t^2) = [(a_1)^2 * t^2] and [(x_1)^2 * t^2] - (h_1 * t^2) = [(b_1)^2 * t^2]. Cancelling the factor of t^2 gives: (x_1)^2 + h_1 = (a_1)^2 and (x_1)^2 - h_1 = (b_1)^2 When no further reduction is possible, we shall say that we have a primitive solution. When this reduction is applied to (8-18), we obtain: x = m^2 + n^2 h = 4*m*n*(m^2 - n^2) The numbers m and n now produce a primitive Pythagorean triple. Thus your solution of 15^2 + 216 = 21^2 15^2 - 216 = 3^2 Is really just a multiple of the primitive solution: 5^2 + 24 = 7^2 5^2 - 24 = 1^2 This corresponds to the first row in the table here: Congruum Problem - MathWorld - Eric Weisstein http://mathworld.wolfram.com/CongruumProblem.html i.e., take m = 2, n = 1. Then x = 5, h = 24, a = 7, b = 1 and as with Pythagorean triples, letting t be a natural number gives an infinite number of solutions. Neat problem! I hope this is clear - let me know if I'm skipping steps where I need to be filling in the details. Finally, notice that writing h = 4*m*n*(m-n)*(m+n) is only possible when h is an element of a primitive congruum. That's why your attempts failed. You needed to write h = 4*m*n*(m-n)*(m+n)*t^2 which works easily when h = 216, m = 2, n = 1, t = 3. - Doctor Paul, The Math Forum http://mathforum.org/dr.math/ |
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