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 2004-01-13, 14:40 #1 michael   Dec 2003 Belgium 4116 Posts counting bricks Some non-artistic king in an ancient country had an idea of building a useless brick wall. A wall which would evoluate during his reign. Every year he added some bricks to his wall in a specific fashion. year1: + year2: + ++ year3: + ++ +++ etcetera where + is the symbol used for a brick. So every year the king added a diagonal layer of bricks. Now one day the king died, and his equally non-artistic son continued building this useless brick wall his dad started. If the king reigned 2 years his wall would be: + ++ containing 3 bricks If as addition to that the son reigns 1 year the wall will become + ++ +++ where the son adds a total of 3 bricks to the wall (which contains 6 bricks now) The question now is if there exist other pairs of years as the example (2,1) for which the son's addition of bricks is equal to the amount of bricks the wall contained when his father died. -michael
 2004-01-14, 00:57 #2 cheesehead     "Richard B. Woods" Aug 2002 Wisconsin USA 22·3·641 Posts In other words, find solutions to (triangular number) * 2 = (another triangular number).
 2004-01-14, 01:50 #3 Wacky     Jun 2003 The Texas Hill Country 32×112 Posts The mersennaries killed them both just after the old king thought up this silly scheme. (0,0)
 2004-01-14, 06:39 #4 michael   Dec 2003 Belgium 5×13 Posts Cheesehead: correct observation Wacky: That is indeed a correct (trivial) solution, but for now let's assume these silly people actually did reign. -michael
 2004-01-14, 11:32 #5 Wacky     Jun 2003 The Texas Hill Country 21018 Posts If I had not thought that the answer was "unacceptable", I would have hidden it.
 2004-01-14, 14:52 #6 wblipp     "William" May 2003 New Haven 93916 Posts Any odd solution to x^2 - 2y^2 = -1 will give solutions ((x-1)/2, (y-x)/2). The simplest way to generate such solutions is to start with the given solution 7^2 - 2*5^2 = -1 and to multiply by 3^2 - 2*2^2 = +1 Using the rule (a^2 - 2b^2)(x^2 - 2y^2) = (ax+2by)^2 - 2(ay+bx)^2 Reparameterizing the relationship into the form of the answer, we get that if (a,b) is a solution, then so is (5a+2b+2,2a+b+1) The first few solutions in this series are (2, 1) (14, 6) (84, 35) (492, 204) (2870, 1189) (16730, 6930) William
 2004-01-14, 15:32 #7 michael   Dec 2003 Belgium 5·13 Posts Correct William, also some math in the ring Z/sqrt(2)Z can show that these are the only solutions. I used slightly different parameters in my solution. In the end i get my solutions from (7+5*sqrt(2))*(3+2*sqrt(2))^k (k=1,2,3,...). Take m,n as the coefficients of the numbers in the ring Z/sqrt(2)Z: (m+n*sqrt(2)) then the first few terms are: (7,5) (41,29) (239,169) (1393,985) Same as for your solution now the results follow as a=(m-1)/2 and b=(n-1)/2 Where my paremeter a is the total amount of years, unlike yours which takes the amount of years reigned by the son. Hope you had as much fun as i did soluting this riddle. -michael
 2004-01-14, 16:27 #9 michael   Dec 2003 Belgium 5·13 Posts If i were a teacher all the points would go to correctness of the proposed solution. I have had trouble with professors myself, sometimes they didn't even want to start reading my solution, reason: not the same as in my course! Correct=Correct to me. -michael

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