Ordered pair satisfying the equation $x^2 + 6x + y^2 = 4$ How many ordered pairs of integers $(x,y)$ satisfy the equation $x^2+6x + y^2 = 4$?
I can't approach this problem. Please help me.

By completing the square we get $(x+3)^2+y^2=13$.
If $x$ and $y$ are integers, so are $x+3$ and $y$ so we require two square numbers that add to $13$. The only option is $9$ and $4$. We can choose which is $9$ and which is $4$ as well as whether we use the positive or negative square roots. Hence $8$ possibilities.
