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 August 2nd, 2010, 09:56 AM #1 Newbie   Joined: Jan 2010 Posts: 20 Thanks: 0 diophantine equation How to identify diophantine equations and how is it solved? Please solve along with an example. Thank you.
 August 2nd, 2010, 12:31 PM #2 Global Moderator   Joined: May 2007 Posts: 6,540 Thanks: 591 Re: diophantine equation http://en.wikipedia.org/wiki/Diophantine_equation Look at the above.
 August 2nd, 2010, 12:44 PM #3 Member   Joined: Aug 2010 Posts: 31 Thanks: 0 Re: diophantine equation A diophantine equation is one where the only allowed solutions are integers. For example: In a pet shop, rats cost 5 dollars, guppies cost 3 dollars and crickets cost 10 cents. One hundred animals are sold, and the total receipts are 100 dollars. How many rats, guppies and crickets were sold? You get the equations: r + g + c = 100 5r + 3g + 0.1c = 100 This simplifies to the diophantine equation: 49r + 29g = 900 This is one equation in two unknowns and it normally has an infinite number of solutions. However, if you further restrict the problem so that c, r and g must be positive integers, then the solution can be found as follows: The number of crickets sold must be a multiple of 10. Try the different values of c: 0, 10, 20, 30, ....., 100. If c = 10, then r + g = 90. Try different values of r. For r=1, r=2, and so on, there are no integer solutions when c=10. Continue trying c=20, 30, ..... When you get to c=80, then r + g = 20, and you eventually find that c=80, r=16, and g=4 is a solution where c, r and g are integers. Really smart people are working on the theory of diophantine equations, but I think it is safe to say that finding solutions to a diophantine equation problem can be described as solving a puzzle.
 August 3rd, 2010, 01:41 AM #4 Math Team   Joined: Apr 2010 Posts: 2,778 Thanks: 361 Re: diophantine equation Can't such an equations, $r + g + c= 100\\ 5r + 3g + 0.1c = 100$ be solved by: $(r+g+c)-(5r+3g+0.1c)=(100-100)=0$ ... $9c=40r+20g$ ... Backsubstitute: $r+g+\frac{40r+20g}{9}=100=\frac{245r+145g}{45}$ and $5r+3g+\frac{40r+20g}{90}=100=\frac{245r+145g}{45}$ ... $g=\frac{4500-245r}{145}=\frac{900-49r}{29}$ This is bizarre, I get the same, as well by backsubstituting the last fraction. Can anyone show me the error, or is the method wrong? Hoempa
 August 3rd, 2010, 03:54 AM #5 Global Moderator   Joined: Dec 2006 Posts: 19,184 Thanks: 1647 It's not bizarre. r + g + c = 100 50r + 30g + c = 1000 Subtracting the first equation from the second gives 49r + 29g = 900, so g = (900 - 49r)/29. You correctly obtained the same result by a longer route.
 August 3rd, 2010, 09:05 AM #6 Math Team   Joined: Apr 2010 Posts: 2,778 Thanks: 361 Re: diophantine equation And so did followingmycoffeecup, with that method.. Thanks, skipjack Hoempa
 August 4th, 2010, 05:17 AM #7 Newbie   Joined: Jan 2010 Posts: 20 Thanks: 0 Re: diophantine equation Thnx for the example. But I still didn't get how to IDENTIFY a diophantine equation.
 August 5th, 2010, 08:21 AM #8 Member   Joined: Aug 2010 Posts: 31 Thanks: 0 Re: diophantine equation A diophantine equation is an equation where the roots must be integers. It depends on the problem that you are solving. In the example above the equations are: r + g + c = 100 50r + 30g + c = 1000 You have two equations in three unknowns and there are an infinite number of solutions. If you restrict r, g, c to being positive integers, then it is diophantine. You can't buy half a rat. If it was a butcher shop instead of a pet store, then it would not be diophantine.

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