Question

How can I solve 3*x^2+y^2=300? (Diophantine equation)

Answer 1

Since you said it is a Diophantine version of the equation, you are looking for solutions with integer values of x and y. x yould have to be 10 or less and y would have to be 17 or less, to make the right side equal to 300. I tried x values of 10 and less and see if there are any y solutions that are integers.

The only integer solution that I could find are (10,0) and (5,15).

3x^2+y^2=300 is the equation of an ellipse, by the way.


See http://en.wikipedia.org/wiki/Diophantine_equation

Answer 2

By definition, we are seeking integer solutions.

Another solution path:

1--3x^2 + y^2 = 300
2--Let x = a and y = ma
3--3a^2 + m^2 a^2 = 300
4--a^2(3 + m^2) = 300
5--a^2 = 300/(3 + m^2)
6--a = sqrt[300/(3 + m^2)]
7--m.....0.....1.....2.....3
...a....10.....-.....-.....5
...x....10.....-.....-.....5
...y.....0.....-.....-....15

Thus, there are only two solutions, (10,0) and (5,15)

Answer 3

Step 7: How do you know that you stop at 3?