The number of ordered pairs of integer $(x,y)$ satisfying equation : $x^2+6x+y^2=4$

Question

I have solved it completing square i.e., $(x+3)^2 + y^2= 13$ form, but couldn't understand how to find the number of ordered pairs.

Answer 1

You are almost there. To write $13$ as sum of two perfect squares, these squares must of course be $\le 13$, hence can only be $0,1,4,9$. By trial and error, we find that only $4+9=13$ and $9+4=13$ are solutions. In the first case, $x+3=\pm 2$, $y=\pm 3$, in the second case $x+3=\pm 3$, $y=\pm 2$. From this you can list a total of eight solutions.

Answer 2

There are not many of them. You can start by finding a pair of squares that sum $13$.

This is a representation of the solution.