# Prove quadratic diophantine has no solutions?

I am trying to prove a quadratic diophantine equation has no integer solutions. Any input would be great, I am interested in the general method for this type of equation so any explanation / link to additional resources would help. Thanks a bunch! The equation always has this form:
$A^2x^2-C^2y^2=Ey-Dx-F,(\{A,C,D,E,F\}>0)$

• For the general theory of quadratic forms over $\Bbb Q$ a standard reference is Part I in J.-P-Serre's A course in Arithmetic, Springer GTM series n.7 – AdLibitum Jul 18 '15 at 12:54
• This can help this formula. artofproblemsolving.com/community/c3046h1049910___4 – individ Jul 18 '15 at 17:28
• Thanks individ, I looked at your link and looks very interesting. The only problem I found is that the value 'k' in the description is not an integer in the following diophantine equation with solutions (0,4): 1296*x^2 - 9*y^2 + 1498*x - 13*y + 196 = 0 – Latin PMB Jul 18 '15 at 23:50
• This formula shows that if the difference of squares, then the solutions to the equation are infinitely many will not. If the decision is their final number. – individ Jul 19 '15 at 7:24
• I can't seem to explain. If the root was intact then the solution is infinitely many. But solutions are not infinitely many. This means that the formula should not give solutions. After conversion, we come to the Pell equation of the form $x^2-t^2y^2=1$ . In this equation except no trivial solutions. Formula you can write only in the case where infinitely many solutions - if no solutions or finite number formula impossible to write. – individ Jul 20 '15 at 4:48

The equation $$a^2x^2-c^2y^2=ey-dx-f$$ is a special case of a quadratic Diophantine equation.