# Christmass Puzzle

Discussion in 'Physics & Math' started by ProCop, Dec 25, 2003.

1. ### DinosaurRational SkepticValued Senior Member

Messages:
4,885
NanoTec: How did you find those values? It would have taken me forever just trying values for Blue, which is what I did to find the first three solutions.

The Reds are both odd & even, but so far the Blues are all odd.

3. ### NanoTecRegistered Senior Member

Messages:
30
An Excel spreadsheet which calculated Red from values of Blue (as Red>Blue) then performed a Int(Red)=Red test. Find TRUE located the answers. MathCAD would be great to have.

For larger numbers I was trying to use a Pascal program I wrote
T:=1149850;
B := 257115;
repeat
R := 1+20*(B-1)*B;
While (T*T)<R do
T:=T+1;
If T*T=R then
If ((T+1) mod 2)=0 then
Writeln('Blue: ',B,' Red:',(T+1) div 2);
B:= B+1;
until B<0;

One had to use a 64-bit integer :m: for it to finally output anything larger than what I first posted.
(257115,574926) (673135,1505175) (4613733,10316620) (12078909,27009261) (82790071,185124226) (216747219,484661515)

5. ### metacristiRegistered Senior Member

Messages:
92
The analysis of the sequences of A yielded nothing (at least this is what I obtained).All I could establish is that the terms having the same index in one sequence are 5 times higher than those in the other sequence,the closed formulas being (n<sup>2</sup>+n) and 5*(k<sup>2</sup>+k) with k,n=1,2,3,4,5..........

So that the shortest approach to the solution of the problem is to solve the Diophantine equation r(r-1)=5b(b-1) where the notation is obvious (r=the number of RED balls,b=the number of BLUE balls).Of course I have never solved a Diophantine equation of the second order so that I tried my luck with the Net.And my expectations have not been betrayed,there are online calculators which solve this type of equations.See http://www.alpertron.com.ar/QUAD.HTM

First the equation need to be re-written (for compatibility reasons).If r=x and b=y --->

x<sup>2</sup>-5y<sup>2</sup>-x+5y=0 (1)

The computer did the work (it is even explained in detail the solution) and came out with:

x<sub>n+1</sub>=9x<sub>n</sub>+20y<sub>n</sub>-14

y<sub>n+1</sub>=4x<sub>n</sub>+9y<sub>n</sub>-6

We have already established that the set (x<sub>1</sub>=6,y<sub>1</sub>=3) and (x<sub>2</sub>=15,y<sub>2</sub>=7) are the first two acceptable sets.Plugging these into the above equations we obtain:

x<sub>3</sub>=100
y<sub>3</sub>=45

x<sub>4</sub>=261
y<sub>4</sub>=117

A clear evidence that there is no constraint that the number of red balls must be even.I suspect that there is no constraint for the number of blue balls to be odd either.

x<sub>5</sub>=1786
y<sub>5</sub>=799

and so on.

It seems that there is an infinite number of solutions.

Last edited: Dec 27, 2003

7. ### DinosaurRational SkepticValued Senior Member

Messages:
4,885
MetaCristi: Thnax for the URL. Looks like a handy site to know.

8. ### metacristiRegistered Senior Member

Messages:
92
You are welcomed Dinosaur,very interesting indeed.I checked many of the links,some of them are really great.

I think there is a logical justification for the conclusion that b (or y)=the number of BLUE balls must be odd.There are possible four cases for x<sub>n</sub> and y<sub>n</sub>:

1.(x<sub>n</sub>=odd,y<sub>n</sub>=odd)
2.(x<sub>n</sub>=odd,y<sub>n</sub>=even)
3.(x<sub>n</sub>=even,y<sub>n</sub>=even)
4.(x<sub>n</sub>=even,y<sub>n</sub>=odd)

CASE1.

x<sub>n</sub>=odd
y<sub>n</sub>=odd

4*x<sub>n</sub>=even
9*y<sub>n</sub>=odd
y<sub>n+1</sub>=even+odd-even=odd

9*x<sub>n</sub>=odd
20*y<sub>n</sub>=even
y<sub>n+1</sub>=odd+even-even=odd

Using the same reasoning for the remaining cases it can be drawn the conclusion that when x<sub>n</sub> is odd x<sub>n+1</sub> must be odd and when x<sub>n</sub> is even x<sub>n+1</sub> must be even also.The same pattern is valid for the set [y<sub>n</sub>,y<sub>n+1</sub>].Since in the two initial independent solutions (6,3) and (15,7) y is always odd it follows from the above considerations that y remains odd for all values of the index (n) whilst x will vary upon the initial value of x plugged in the recurrence equations.

9. ### NanoTecRegistered Senior Member

Messages:
30
<b>metacristi</b> Thanks for the answer, and especially the explanation.