20040113, 14:40  #1 
Dec 2003
Belgium
5×13 Posts 
counting bricks
Some nonartistic 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 nonartistic 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 
20040114, 00:57  #2 
"Richard B. Woods"
Aug 2002
Wisconsin USA
2^{2}·3·641 Posts 
In other words, find solutions to (triangular number) * 2 = (another triangular number).

20040114, 01:50  #3 
Jun 2003
The Texas Hill Country
3^{2}×11^{2} Posts 
The mersennaries killed them both just after the old king thought up this silly scheme. (0,0)

20040114, 06:39  #4 
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 
20040114, 11:32  #5 
Jun 2003
The Texas Hill Country
3^{2}×11^{2} Posts 
If I had not thought that the answer was "unacceptable", I would have hidden it.

20040114, 14:52  #6 
"William"
May 2003
New Haven
2^{3}×5×59 Posts 
Any odd solution to x^2  2y^2 = 1 will give solutions ((x1)/2, (yx)/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 
20040114, 15:32  #7 
Dec 2003
Belgium
65_{10} 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=(m1)/2 and b=(n1)/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 
20040114, 16:21  #8 
"William"
May 2003
New Haven
2^{3}×5×59 Posts 
I wasn't sure I had all the answers. I had figured out it was all, but you posted that information first.
I had to check if powers of (3^22*2^2) are all the solutions of the Pell Equation 2x^2+1=y^2 My solutions can be transformed into your solutions. By expressing the the quadratic multiplication as a matrix, then expressing the series in terms of the eigenvalues and eigenvectors. Yes, it was fun. But do you really think you're playing fair? You asked for the answers in the form (2,1)  father years, son years. It's much easier to find answers in the form (2,3)  father years, total years. We both did that. But about half the effort (for me) was translating that answer into the requested format. You left that part as a exercise for the interested reader. I've watched my teenager loose points in math competitions and SATs for having solved the problem but failing to cast the answer in the requested terms  perhaps my viewpoint is only the result of being father to such a teenager. William 
20040114, 16:27  #9 
Dec 2003
Belgium
101_{8} 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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