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.
 A: 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.
A: Hint: You can "complete the square" - the object of which is to get rid of the $6x$ term which is awkward.
$$x^2+6x+y^2=x^2+6x+9-9+y^2=(x+3)^2+y^2-9$$
This transforms your original equation into the sum of two squares equals a positive integer. You then have only a few cases to try (remembering that integers can be negative).
A: If you look at $f(x) = 4-x^2-6x$ you will see that $f(x) \ge 0 $ iff $x \in
[-3 -\sqrt{13}, -3+\sqrt{13}]$.
There are only a few integer $x$ values to try $\{-6,...,0\}$.
Any solution is of the form $y^2 = f(x)$.
A: For all integer $x\gt0$, there is no $y$ that satisfy $y^2=-x^2-6x+4$. So, the only solutions are $(0, 2)$ and $(0, -2)$. For integer $x\lt0$, do the same and look for which $x$, the quadratic equation $-x^2-6x+4\ge0$ which are $x\in\{-6,-5,-4,-3,-2,-1\}$. Find $y$ now.
