Relay-Version: version B 2.10 5/3/83; site utzoo.UUCP Posting-Version: version B 2.10.2 9/18/84; site faron.UUCP Path: utzoo!linus!faron!bs From: bs@faron.UUCP (Robert D. Silverman) Newsgroups: net.math Subject: Re: A real problem! (Not a polar bear problem) Message-ID: <383@faron.UUCP> Date: Fri, 15-Nov-85 10:35:36 EST Article-I.D.: faron.383 Posted: Fri Nov 15 10:35:36 1985 Date-Received: Sat, 16-Nov-85 03:37:31 EST References: <509@klipper.UUCP> <1096@jhunix.UUCP> <2081@umcp-cs.UUCP> <775@mmintl.UUCP> <45@yale.ARPA> Distribution: net Organization: The MITRE Coporation, Bedford, MA Lines: 92 > > Enough silly polar bear problems. > Here's a problem that requires a little more sophistication: > > The 3-4-5 triangle has integral area (area=6). > The 13-14-15 triangle has integral area (area=84). > Find all triangles with sides n-(n+1)-(n+2) that have > integral area. > > Note: This isn't an elementary problem. > I'll send the answer to anybody who wants it, and maybe even post > it . . . > -- > Thomas Andrews > andrews-thomas@yale Not to contradict you but the solution IS elementary. One need only know a little elementary number theory including simple diophantine equations. The area of a triangle is: SQRT( s (s-a) (s-b) (s-c)) where a,b,c are the lengths of the sides and s is the semi-perimeter. (Heron's formula) Substitute n, n+1, n+2 and we obtain: 1/4 (n+1) sqrt( 3 (n+3) (n-1) ) Thus we require that 3(n+3)(n-1) be a square, say A^2. Now it is easy to see that either n+3 or n-1 must be divisible by 3, so n = 0 or 1 mod 3. Further, A must be an multiple of 3, say 3h. ------------------------------------------------------------------------------ Case 1: n = 1 mod 3 Let n = 3 k + 1 for some k we obtain: k(3k+4) = h^2 Solving for k we obtain via the quadratic formula: (-4 +/- sqrt(16 + 12h^2))/6 And we require that 4 + 3h^2 be a square ------------------------------------------------------------------------ Case 2: n = 0 mod 3 Let n = 3 k we obtain: (k+1)(3k-1) = h^2 again we obtain 4 + 3h^2 must be a square by solving for k. ----------------------------------------------------------------------- Combining these we find: k = (sqrt(4 + 3h^2) -1) / 3 or k = (sqrt(4 + 3h^2) - 2) /3 We now must have 4 + 3h^2 = s^2 or s^2 - 3h^2 = 4 This is a (modified) Pells equation. Its solution is given by: P (1/2 (s + sqrt(3) h ) P = 0, 1, 2, ... 0 0 where s and h are the least positive solution. (4,2) 0 0 One can also obtain new solutions via the linear recurrence relation: s = 4s - s i i-1 i-2 This yields the sequence of solutions: s = (2, 4, 14, 52, 214, ...) h = (0, 2, 8, 30, 112, ...) Note: One can derive the solution to the above by elementary means using a little theory about recurrence relations OR if you know some algebraic number theory by finding units in the quadratic extension field Q(sqrt(-3)). If we substitute the values given for h, we can solve for k and hence get n. This solution is essentially complete. Bob Silverman (they call me Mr. 9)