x^2-6xy+3y^2=100 x,y are natural numbers.
Help would be appreciated.
Wolfram Alpha gives a complete set of integer solutions. They look quite messy and I haven't checked them out in any way.
The first few though are not difficult following the quadratic/discriminant route.
You need $\displaystyle 6y^{2}+100$ to be a square, so check out the sequence $\displaystyle 6y^{2}=21, 44, 69, 96,...$ .
$\displaystyle 6y^{2}=96$ gets you an early hit and a second comes not much further on.
I may have been using a more complicated system than you. (I had to define a z that was equal to the square root of the discriminant.) This is a rather common difficulty of mine.
How did you code your question to Wolfram|Alpha? I tried to do that to check my work but was unable to get it to solve the problem
-Dan
I'm not a regular user of Wolfram Alpha, in fact this is the first time that I've actually asked it a question.
I simply typed in the equation and pressed the compute button.
It came back with a whole pile of stuff, including a graph, which was pleasing to see, and a section headed 'Integer Solutions '.