Finding integer solutions

Question

Find all integer solutions to the problem $y^2+x^2-6x=0$. How I solved this was to complete the square then finding the coordinates: $(0,0), (6,0), (3,3), (3,-3)$. What I would like to know is there another method to solving this kind of equation (finding the integer solutions) other than completing the square and drawing the circle? If so please could you explain it to me. Thanks.

Answer 1

HINT:

$$(y)^2+(x-3)^2=3^2\implies y^2=\cdots\le9\iff-3\le y\le3$$

Clearly, if $y=a$ is a solution, $y=-a$ will also be.

So, we can limit the test for $0\le y\le3$

Answer 2

From the given, we have $(6-x)x = y^2 \ge 0$

Case-1: $6 – x \le 0$ and $x \le 0$. This gives an impossible result.

Case-2: $6 – x \ge 0$ and $x \ge 0$. This gives $0 \le x \le 6$.

This limits the search area for integral solutions for x and y.