# Trivial case for Pell's equation

I was reading Pell's equation, and in the beginning there is a statement:

Notice that if $D = d^2$ is a perfect square, then this problem can be solved using difference of squares. We would have $x^2 - Dy^2 = (x+dy)(x-dy) = 1$, from which we can use casework to quickly determine the solutions.

How to determine the solution? my thoughts: $$(x+dy)= {1\over(x-dy)}$$ and since x+dy is an integer($x,y,d \in \Bbb Z$), we have $$x-dy = \pm1 = x+dy$$ $$2dy = \pm1$$ $$y = \pm{1\over2d}$$

I guess I broke it down into two cases ($\pm1$) but somehow I got there does not exist any solution for the trivial case (x,y must be integral solutions) but x = 1 y = 0 is a solution, where did I make a mistake? does this trivial case have infinitely many solutions?

• Should be $2dy=0$! – bilaterus Nov 16 '15 at 18:29
• $x-dy=x+dy$ means $2dy=\pm 1$? – Thomas Andrews Nov 16 '15 at 18:36
• ...what a stupid mistake – watashiSHUN Nov 16 '15 at 21:12

$$(x+dy)(x-dy) = 1$$ implies that we have two cases: $$x+dy = x-dy = 1,$$ or $$x+dy = x-dy = -1.$$ In the first case, adding the equations together gives $$2x = 2, \quad x = 1,$$ from which we must have $y = 0$, and in the second, $$2x = -2, \quad x = -1,$$ which also implies $y = 0$. These solutions are unique for nonzero integers $d$.
Your algebra is not correct because of the way you wrote your equations. Let's take the positive case: you wrote $$x-dy = 1 = x+dy.$$ Then I presume you "moved" the LHS terms to the right; i.e., you subtracted $x$ and added $dy$ to the LHS and the RHS. But if you do this, you have to do it to the middle also, otherwise you would get the nonsensical $$0 = 1 = 2dy,$$ when in truth, you should have written $$0 = 1 - x + dy = 2dy.$$ When you write compound equalities, any manipulation must be done to all components.
The equation $$x^2 - d^2y^2 = 1$$ says that we have two square numbers one after the other, $0^1$ and $1^2$ is an example - there are no others since squares get further and further apart.
Proof: For $m^2 = 1 + n^2$ with $n$ positive we must have $m > n$ but $m = n+1$ is too large (since $(n+1)^2 = n^2 + 2n + 1$) and so is any larger $m$ so there are no solutions.