# How to study results of Diophantine equation?

I finally managed to learn a bit of number theory and Diophantine equation(with the help of Arturo Magidin's great answer in what type of math is this?). But I'm wondering what's the next step after?

How do you take all the results and figure out which one is correct? If I use a small sum and very specific numbers then I may only get a handful of results but as my numbers and variables get bigger I get many many results(I stop my computer after a million). I'm new to math to I thought I'd ask the community here what are the approaches to this step now?

I've been thinking and, correct me if I'm wrong, but this is probably a probability question now since every answer is technically correct. I know the correct answer is somewhere in the millions of results I get, but I'm trying to at least narrow it down.

Do I use a computer program to study each result against parameters I set(i.e. no result can change more than 5% and only 2 colors can be chosen at a time), or is there a mathematical way of solving it?

Thanks in advance! (sorry if I'm wording this question incorrectly..I'm not 100% sure the name of what I am searching for)

Here's some more information to clear it up more. For example I have the following:

color  value   quantity
red       20    2
blue    5   8
green   10  2

total       100


If only the value and the total is given, I will find there is 36 possible answers:

#1 Found : 20.0*0.0 red + 5.0*0.0 blue + 10.0*10.0 green = 100.0
#2 Found : 20.0*0.0 red + 5.0*2.0 blue + 10.0*9.0 green = 100.0
#3 Found : 20.0*0.0 red + 5.0*4.0 blue + 10.0*8.0 green = 100.0
#4 Found : 20.0*0.0 red + 5.0*6.0 blue + 10.0*7.0 green = 100.0
#5 Found : 20.0*0.0 red + 5.0*8.0 blue + 10.0*6.0 green = 100.0
#6 Found : 20.0*0.0 red + 5.0*10.0 blue + 10.0*5.0 green = 100.0
#7 Found : 20.0*0.0 red + 5.0*12.0 blue + 10.0*4.0 green = 100.0
#8 Found : 20.0*0.0 red + 5.0*14.0 blue + 10.0*3.0 green = 100.0
#9 Found : 20.0*0.0 red + 5.0*16.0 blue + 10.0*2.0 green = 100.0
#10 Found : 20.0*0.0 red + 5.0*18.0 blue + 10.0*1.0 green = 100.0
#11 Found : 20.0*0.0 red + 5.0*20.0 blue + 10.0*0.0 green = 100.0
#12 Found : 20.0*1.0 red + 5.0*0.0 blue + 10.0*8.0 green = 100.0
#13 Found : 20.0*1.0 red + 5.0*2.0 blue + 10.0*7.0 green = 100.0
#14 Found : 20.0*1.0 red + 5.0*4.0 blue + 10.0*6.0 green = 100.0
#15 Found : 20.0*1.0 red + 5.0*6.0 blue + 10.0*5.0 green = 100.0
#16 Found : 20.0*1.0 red + 5.0*8.0 blue + 10.0*4.0 green = 100.0
#17 Found : 20.0*1.0 red + 5.0*10.0 blue + 10.0*3.0 green = 100.0
#18 Found : 20.0*1.0 red + 5.0*12.0 blue + 10.0*2.0 green = 100.0
#19 Found : 20.0*1.0 red + 5.0*14.0 blue + 10.0*1.0 green = 100.0
#20 Found : 20.0*1.0 red + 5.0*16.0 blue + 10.0*0.0 green = 100.0
#21 Found : 20.0*2.0 red + 5.0*0.0 blue + 10.0*6.0 green = 100.0
#22 Found : 20.0*2.0 red + 5.0*2.0 blue + 10.0*5.0 green = 100.0
#23 Found : 20.0*2.0 red + 5.0*4.0 blue + 10.0*4.0 green = 100.0
#24 Found : 20.0*2.0 red + 5.0*6.0 blue + 10.0*3.0 green = 100.0
#25 Found : 20.0*2.0 red + 5.0*8.0 blue + 10.0*2.0 green = 100.0
#26 Found : 20.0*2.0 red + 5.0*10.0 blue + 10.0*1.0 green = 100.0
#27 Found : 20.0*2.0 red + 5.0*12.0 blue + 10.0*0.0 green = 100.0
#28 Found : 20.0*3.0 red + 5.0*0.0 blue + 10.0*4.0 green = 100.0
#29 Found : 20.0*3.0 red + 5.0*2.0 blue + 10.0*3.0 green = 100.0
#30 Found : 20.0*3.0 red + 5.0*4.0 blue + 10.0*2.0 green = 100.0
#31 Found : 20.0*3.0 red + 5.0*6.0 blue + 10.0*1.0 green = 100.0
#32 Found : 20.0*3.0 red + 5.0*8.0 blue + 10.0*0.0 green = 100.0
#33 Found : 20.0*4.0 red + 5.0*0.0 blue + 10.0*2.0 green = 100.0
#34 Found : 20.0*4.0 red + 5.0*2.0 blue + 10.0*1.0 green = 100.0
#35 Found : 20.0*4.0 red + 5.0*4.0 blue + 10.0*0.0 green = 100.0
#36 Found : 20.0*5.0 red + 5.0*0.0 blue + 10.0*0.0 green = 100.0


As you can see, in the possibilities I get the correct answer but many other answers also. Now say I add one more red(so the total red is 3) then I now have 49 results, but some of the results in second set are not likely if you factor in the relationship with the first result set. I assume as I get more data results, I can more accurately remove the results that don't work. I'm not sure if there's a branch of math that deals with this?

Update: Is this probability? fitness(or cost) functions? mathematical optimization? or is this even a math problem?

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I've read your other questions and I think you're requiring that solutions be positive integers. Generally, if a system is solvable and negative integers are allowed there are an infinite number of solutions, however your restriction means you can only shuffle the calories around so much. So you should be able to compute all of the solutions (this could be quite large!) and all of them are allowable. If this was posed as a guessing game then I suppose the probability of guessing the correct numbers out of a pool is one over n, with n=solutions i.e. n_1 = {x1,x2,x3,...}. My thoughts. – ttt Jun 6 '11 at 23:22
Thanks Tony! Your 100% correct, my problem is now that the results are very large(some of the tests I am running is over 1000 variables), and I'm trying to figure out how to narrow it down. Even with the sum changings and values changing within intervals, there must be a way to figure out which ones are most likely true. For example, if calories change by 5 then all the values that changed to 10 from the previous answer should not be allowed..for example 5x + 10y +20z = 25 would give me all results possible but if the next interval the values changed to 5x + 10y + 25z =25 – Lostsoul Jun 7 '11 at 0:41
then some of the previous answers could be eliminated from the last interval. I'm not sure if its a math problem or a computer problem in studying these relationships..I am working on doing some reading on bayes probability theorem, but i'm not sure. – Lostsoul Jun 7 '11 at 0:43
I'm going to steer you away from probability because the question I think you are asking really depends on the specific diophantine equation(s) in question, that is how the coefficients and total are related by division and common factors. Generally, a diophantine equation has a solution in the integers when the greatest common divisor of the coefficients divides the total. It might be helpful to edit your question to refine what exactly you want answered. – ttt Jun 7 '11 at 4:08
In your example it's useful to note relationships like, "y is worth twice as much as x", so any solution yields another solution where y is incremented by one and x is decremented by two. In that case you can fix z=0 and look the number of ways 5 and 10 can form 25. Equivalently, how 1 and 2 can form 5. – ttt Jun 7 '11 at 4:12

@Lostsoul, good luck. Here's an experiment you might try. Consider the equation $23x+29y=10000$. This has plenty of solutions in positive integers $x,y$, spaced out uniformly on a line. Now look at $23x+29y=10001$. This also has lots of solutions, spaced out uniformly, and fairly distant from the solutions of the 10000 equation - they won't help you at all to pick out a solution of that first equation. Then look at $23x+29y=10002$, and $23x+29y=10003$, and $23x+29y=10004$. You'll find the same thing happening, but, really, do the experiment and see for yourself what you're up against. – Gerry Myerson Jun 8 '11 at 0:24