# How to solve this eqution

aravindsubramanian
I need to solve this kind of equqtions

$$x^2-y^2+ax-by = c$$

a,b,c are know.I want to find out x & y.

note:
a,b & c are positive integers

I need only positive x & y pairs in integers.

Last edited:

## Answers and Replies

Science Advisor
Homework Helper
Complete the square for both x and y and you will see that the set of points satisfying the equation form either an ellipse or hyperbola depending on the values of a, b and c.If you need to you can compare the equation with one of the standard forms to determine foci, major & minor radii and so forth.

In any case, you can simply solve the given equation or a simplified version of it for either x or y in terms of the other variable and substitute values of one into the solution to find the corresponding values of the other variable.

ramsey2879
Tide said:
Complete the square for both x and y and you will see that the set of points satisfying the equation form either an ellipse or hyperbola depending on the values of a, b and c.If you need to you can compare the equation with one of the standard forms to determine foci, major & minor radii and so forth.

In any case, you can simply solve the given equation or a simplified version of it for either x or y in terms of the other variable and substitute values of one into the solution to find the corresponding values of the other variable.
Thus we have
(x+a/2)^2 -(y+b/2)^2 = c + (a^2-b^2)/4
(x+a/2-y-b/2)*(x+a/2+y+b/2) = c + (a^2 -b^2)/4
These two factors differ by 2y+b
To reduce typing let s= x+a/2-y-b/2 and t = x+a/2+y+b/2
then t-s=2y+b

Case 1 = b and a are both odd. Then the two factors must be of opposite parity, i.e. x and y are of opposite parity. Let P = the odd part of c+(a^2-b^2)/4 then c+(a^2-b^2) = 2^k*P where k>1. From Euclid's lemma and the fact that the above factors are both integers of opposite parity, there are only two possibilities.
1a s = 2^k*N where N divides P and t = P/N or
2a t = 2^k*N where N divides P and s = P/N
We need only to factor P. As an example, let P = 15 = 1*3*5 then for each of 1a and 2a there are 4 possibilities that have to be eliminated: N=1,3,5,or 15. With each choice y = (t-s-b)/2 from which x can be calculated and we can varify whether or not x and y are both positive.

I think the case where a and b are both even is similar to case 1. However, s and t, i.e. x and y, must be of the same parity

Case 2, a and b are of opposite parity
Multiply both sides by 4 to get
(2x+a-2y-b)*(2x+a+2y+b) = 4c+a^2-b^2 = s*t
here t-s = 4y-2b

This can be rewritten as s*t = P where P is odd, Factor P as N and P/N. For instance let s= 1,3,5,or 15 and t equal 15,5,3,1 respectively for P = 15. The solution continues as for case 1: first solving y = (t-s-2b)/4 and then solving t= 2x+a+2y+b for x.