MHB Show that the diophantine equation has only these solutions

[evinda]

Hello! (Wave)

I want to show that the diophantine equation $x^2+y^2=2$ has only these solutions:
$$(x,y)=(1,1),(1,-1), (-1,1), (-1,-1)$$

That's what I have tried:

$$x^2+y^2=2 \Rightarrow x^2=2-y^2$$

$$x^2>0 \Rightarrow 2-y^2>0 \Rightarrow y^2<2 \Rightarrow -\sqrt{2}<y< \sqrt{2} \Rightarrow y=\pm 1,0$$

• $y=0: x= \pm \sqrt{2} \notin \mathbb{Z}$
• $y=\pm 1: x= \pm 1$

So, the solutions are $(\pm 1, \pm 1), (\pm 2,0)$.

Is it right? Could I also find the solutions, using modulo?

Also, how can I show that these solutions are the only ones? (Thinking)

 

[mathbalarka]

Are you sure you are asked to find integer solutions to $x^2 + y^2 = 2$? Is there any chance *rational* solutions were asked?

In the case of integers, it's trivially easy and you're overcomplicating it. $x^2 + y^2 = 2$ and $x^2, y^2$ are both positive (squares can't take negative values). Hence $x^2 \leq 2$ and $y^2 \leq 2$. The only possible integer values of $x$ and $y$ satisfying these are $(x, y) = (\pm 1, \pm 1), (\pm 1, 0), (0, \pm 1), (0, 0)$

And from all of them, only the first four $(\pm 1, \pm 1)$, i.e., $(1, 1), (-1, 1), (1, -1), (-1, -1)$ satisfies the equation. No modulo argument is needed.

EDIT : Oh I observed that you enumerated the wrong solutions $(x, y) = (\pm 2, 0)$. There doesn't work as $x^2 + y^2 = 4 + 0 \neq 2$.