Show that there are only trivial solutions

Question

How can I show that the only solutions of the diophantine equation $x^2+y^2=1$ are the trivial ones: $(x,y)=(0,1), (0,-1), (1,0), (-1,0)$ ?

That's what I thought:

$$x \equiv 0,1 \pmod 2 \Rightarrow x^2 \equiv 0,1 \pmod 2$$ $$y \equiv 0,1 \pmod 2 \Rightarrow y^2 \equiv 0,1 \pmod 2$$

$x^2+y^2 \equiv 0,1 \pmod 2 $

Since $0 \not\equiv 1 $, we want $x^2+y^2 \equiv 1 \pmod 2$ and this relation is satisfied, when $x \equiv 0 \pmod 2 \wedge y \equiv 1 \pmod 2$ and when $x \equiv 1 \pmod 2 \wedge y \equiv 0 \pmod 2$

Is it right so far? If so, how can I continue, in order to show that the only solutions are these one: $(x,y)=(0,1), (0,-1), (1,0), (-1,0)$ ?

Answer 1

You don't need to use mod. Note that $$x^2+y^2=1\Rightarrow 1-x^2=y^2\ge 0\Rightarrow -1\le x\le 1\Rightarrow x=0,\pm 1.$$

For $x=0$, we have $x^2+y^2=1\Rightarrow y^2=1\Rightarrow y=\pm 1.$

For $x=\pm 1$, we have $x^2+y^2=1\Rightarrow 1+y^2=1\Rightarrow y=0.$

Hence, there are only four cases as $$(x,y)=(0,1),(0,-1),(1,0),(-1,0).$$

Answer 2

If a number between $|x|$ and $|y|$ is greater than one, $x^2+y^2$ is greater than one, hence the solutions have to be found in $\mathbb{Z}\times\mathbb{Z}\cap\{(x,y):\max(|x|,|y|)\leq 1\}$.

Answer 3

Maybe is not a number-theoretic answer, but you could see that $x^2+y^2=1$ with $x,y\in\mathbb{R}$ is a circle $\gamma$ of radius one centered at the origin, so every point of coordinates $(x,y)$ on $\gamma$ has distance $1$ from the origin. If you restrict to $(x,y)\in\mathbb{Z}^2$ you see that those are the only solutions.