Solving the diophantine equation ax2+bxy+cy2=n, where b2-4ac > 0 and is not a perfect square, n non-zero, gcd(a,b,c)=1


If gcd(a,n)=1, the algorithm finds representatives with least positive y.
(See paper.) The standard method is due to Gauss (See G.B. Mathews Number Theory, page 97 or note.)

E = 1 is verbose.

Enter a:
Enter b:
Enter c: (gcd(a,b,c)=1)
Enter n (non-zero):
Enter E (0 or 1):

Last modified 27th May 2009
Return to main page