# who know how to solve this nonlinear diophantine equation

Discussion in 'Math Research' started by None, Nov 23, 2006.

1. ### NoneGuest

ax^2+bx = cy^2+dy (a,b,c,d are Integer constants and x,y are
variables)
and this equation has little condition
a+b = c+d

None, Nov 23, 2006

2. ### Gerry MyersonGuest

a x^2 + b x = c y^2 + d y
4 a^2 x^2 + 4 a b x = 4 a c y^2 + 4 a d y
4 a^2 x^2 + 4 a b x + b^2 = 4 a c y^2 + 4 a d y + b^2
(2 a x + b)^2 = 4 a c y^2 + 4 a d y + b^2
u^2 = 4 a c y^2 + 4 a d y + b^2
a c u^2 = 4 a^2 c^2 y^2 + 4 a^2 c d y + 4 a c b^2
a c u^2 + a^2 d^2 = 4 a^2 c^2 y^2 + 4 a^2 c d y + a^2 d^2 + 4 a c b^2
a c u^2 + a^2 d^2 = (2 a c y + a d)^2 + 4 a c b^2
a c u^2 + a^2 d^2 = v^2 + 4 a c b^2
v^2 - a c u^2 = a^2 d^2 - 4 a c b^2

(*) v^2 - D u^2 = k,

where v = 2 a c y + a d,
D = a c,
u = 2 a x + b,
k = a^2 d^2 - 4 a c b^2.

Now there are standard techniques for solving (*).
See Pell's equation.

Gerry Myerson, Nov 23, 2006

3. ### Zbigniew KarnoGuest

Gerry Myerson napisal(a):
In 6th line should be:

a c u^2 = 4 a^2 c^2 y^2 + 4 a^2 c d y + a c b^2

Thus in the equation (*):

k = a^2 d^2 - a c b^2 .

Zbigniew Karno, Nov 27, 2006