I have this Equation :
$xy + y^2 = 19 + 2x^2$
I need to find every $(x , y)$ where $x$ and $y$ are integers.
I first tried to solve it by dividing everything by $y^2$ , but $19$ creates a problem , because we will get term $19/y^2$ , which is cumbersome & may or may not be Integer.
I next tried to solve for $x$ and check when is square root of Discriminant integer, but not sure how to check that.
Can anyone help me out ?
This is Diophantine Equation , involving Quadratic terms.
Let us take the given Equation :
$xy+y^2=19+2x^2\tag{1}$
Move the terms around , to get :
$xy+y^2-2x^2=19$
$xy-x^2+y^2-x^2=19$
We can now factorize this :
$x(y-x)+(y+x)(y-x)=19$
$[x+y+x](y-x)=19\tag{2}$
We have 2 Integers , which give Prime Number $19$ when multiplied.
Hence , the Numbers must be either $\pm 1$ or $\pm 19$.
$[x+y+x]=+19 \land (y-x)=+1\tag{3}$
$[x+y+x]=+1 \land (y-x)=+19\tag{4}$
$[x+y+x]=-19 \land (y-x)=-1\tag{5}$
$[x+y+x]=-1 \land (y-x)=-19\tag{6}$
Solve these
twofour Sets of Equations to get the Solutions , selecting Integer Solutions.With (3) & (4) , We will get Solutions $(6,7)$ & $(-6,13)$ which are Integers.
With (5) & (6) , User TonyK gives the Solutions $(-6,-7)$ & $(6,-13)$ which are Integers too.