I followed one convincing proof in this related post but I don't understand the assumption and conclusion.
The proof goes as follows:
Let $r\in\mathbb{Q}$ such that $\frac{x²+x+\sqrt{2}}{y²+y+\sqrt{2}}=r$ and suppose that $\frac{x}{y}$ is a irreducible fraction. Then $$\frac{x²+x+\sqrt{2}}{y²+y+\sqrt{2}}=r\Leftrightarrow > x²+x+\sqrt{2}=(y²+y+\sqrt{2})r\Leftrightarrow > x²+x-y²r-yr=\sqrt{2}(r-1).$$ If $r\neq 1$ we have $\sqrt{2}=\frac{x²+x-y²r-yr}{r-1}\in\mathbb{Q}$, but this is absurd. So we get $r=1$ and $$x²+x+\sqrt{2}=y²+y+\sqrt{2}\Rightarrow > y(y+1)=x(x+1).$$ If $x=-1$, we have $y(y+1)=0$ and so $y=0=x+1$ or $y=x=-1$. If $x\neq -1$ we have $$x=y(\frac{y+1}{x+1}).$$ If $y=-1$, it is analogous. We can suppose that $y\neq -1$. If $y=0$, $x=0$. Suppose $-1\neq y\neq 0$. We have $$\frac{x}{y}=\frac{y+1}{x+1}. $$Since $\frac{x}{y}$ is irreducible, follows the last equality that $x=y$.
I have two questions:
- Is it valid to suppose that $\frac{x}{y}$ is an irreducible fraction? Wouldn't that ignore other values?
- How assuming $\frac{x}{y}$ is irreductible follows that $x$=$y$ for $\frac{x}{y}=\frac{y+1}{x+1}$?
I am particularly interested in the first question as it conflicts with my developing rigour in Mathematics.
No, it's not allowed to even consider $x$ and $y$ integers.
When you arrive at $$ x(x+1)=y(y+1) $$ you're almost done, because this can be rewritten as $$ x^2-y^2+x-y=0 $$ hence $$ (x-y)(x+y+1)=0 $$