number of ordered pairs of integers $(x,y)$ satisfying the equation

Question

I need to find the number of ordered pairs of integers $(x,y)$ satisfying this equation:

$$x^2 + 6x + y^2 = 4.$$

I have tried, and I think $x<0 . $

Is there a specific way to solve such equations?

Answer 1

$(x+3)^2 + y^2 = 13$. So $y^2 \leq 13$, and $|y| \leq \sqrt{13} \approx 3$. So you can find $y's$ and then $x's$. Note that they are integers so you can round down.

Answer 2

it's an equation of a circle

http://www4b.wolframalpha.com/Calculate/MSP/MSP145620gcidadhe55cg660000300770ba8ae7ghf3?MSPStoreType=image/gif&s=59&w=200.&h=204.&cdf=RangeControl

these are integer answers :

$x=-6 , y=\pm2$

$x=-5 , y=\pm3$

$x=-1 , y=\pm3$

http://www.wolframalpha.com/input/?i=x%5E2%2B6x%2By%5E2%3D4

Answer 3

We have $\displaystyle x^2+6x+y^2-4=0$

As $x$ is integer, so must be real $\implies \Delta=6^2-4(y^2-4)=4(13-y^2)\ge0\iff y^2\le13$