A number theory problem by Subhendu Biswas

$\begin{aligned} x^2+6x+y^2 & = 4 \\ x^2 + 6x + 9 - 9 + y^2 & = 4 \\ (x+3)^2 + y^2 & = 13 \\ \end{aligned}$

There are only a pair of squares that sum up to $13$ : $2^2+3^2 = 13$ . Therefore,

$\begin{cases} x+3 = \pm 2 & \Rightarrow x = -1, - 5 & \Rightarrow \begin{cases} (-1, \pm 3) \\ (-5,\pm 3) \end{cases} \\ x+3 = \pm 3 & \Rightarrow x = 0, - 6 & \Rightarrow \begin{cases} (0, \pm 2) \\ (-6,\pm 2) \end{cases} \end{cases}$

There are $\boxed{8}$ ordered pairs of integer $(x,y)$ that satisfy the equation.

Let us solve for x in terms of y using the Quadratic Formula

$x^2 + 6x + (y^2 - 4) = 0 \Rightarrow x = \frac{-6 \pm \sqrt{6^2 - 4(1)(y^2 - 4)}}{2} = -3 \pm \sqrt{13 - y^2}.$

In order for x to be an integer, we require the discriminant to be a non-negative perfect square that only admits integer values for y. For $13 - y^2 \ge 0$ , we require $y = 0; \pm 1; \pm 2; \pm 3$ . Only $y = \pm 2, \pm 3$ yield perfect square values of these seven choices. Our total allowable set includes $\boxed{8}$ integral pairs:

$(x,y) = (0, \pm 2); (-6, \pm 2); (-1, \pm 3); (-5, \pm 3).$

We may rewrite the expression as the following exp: $(x+3)^2=13-y^2 \rightarrow -3 \le y \le 3$ ] It is very easy to get: $\boxed{8}$ roots