Proof that $x^2 -3y^2 = 1$ has infinite solutions. ( x and y are integers)

Question

I have to explain this to my brother who is in eighth grade and I would really love if you could tell this in simple terms (I'm no Maths guy).

Answer 1

Assuming we want integer solutions, it is easily noted that $(2, 1)$ is a solution. Further, if you have one solution $(u, v)$ to this equation (refer Pell equations link given above for more), we can generate another solution as $(2u+3v, u+2v)$. For a check, see

$$(2u+3v)^2 - 3(u+2v)^2 = u^2-3v^2 = 1$$

Now as the successive terms we generate are strictly greater than the earlier ones, we can generate an infinite number of them!

Answer 2

Take any $y$ you want, you can always find $x = \sqrt{1 + 3y^2}$ so that $x^2 - 3y^2 = 1$. Because you can choose an infinite amount of $x$ values, you can find an infinite amount of (different!) pairs.