$$ \ln (x^2 + y^2) + {2x^2\over x^2 + y^2} = 0$$
I solved the easy way by setting each term equal to zero:
$$ {2x^2\over x^2 + y^2} = 0$$ $$ 2x^2 = 0$$ $$x=0$$
Then: $$\ln (x^2 + y^2)=0$$ $$\ln\, (\,y^2)=0$$ $$y=\pm 1$$
I'm looking for other integer solutions for this problem.
Thank you.
There are no others.
If $x$ is a non-zero integer, and $y$ is any integer, then $x^2+y^2$ is a positive integer, so $$\ln\left(x^2+y^2\right)\ge\ln(1)=0.$$ Thus, since $\frac{2x^2}{x^2+y^2}$ will be positive, $(x,y)$ cannot be a solution to the equation in such a case.
Note that since the equation's left-hand side is undefined (so that the equation fails to hold) when $(x,y)=(0,0).$ Thus, excluding this possibility, the equation becomes equivalent to $$\left(x^2+y^2\right)\ln\left(x^2+y^2\right)+2x^2=0.$$ The graph here plots the solutions to this equation.