$\sqrt{x+1}, \sqrt{x+2}, \sqrt{x+3}$ cannot be all three rational numbers. $x$ is a rational number. Then,
$\sqrt{x+1}, \sqrt{x+2}, \sqrt{x+3}$ cannot be all three rational numbers. Any idea? Thanks!
I tried to solve this problem and I arrived to prove that in natural set (positive integers) we cannot have:
$m^2 + n^2 = k^2$ and
$m^2 + 2n^2 = l^2$.
Also is rather easy to see that $m$ is odd, $n$ is even, also $n$ is multiple of $3$.
But I'm not sure that is the way.

Note that your question is equivalent to showing that for a rational $x$, all of $\sqrt{x-1}$, $\sqrt{x}$ and $\sqrt{x+1}$ cannot be rational. If possible, on the contrary, let that be true.
Let $x=\frac{a^2}{b^2}$ with $(a,b)=1$ so that
\begin{align}
\sqrt{x-1}&=\frac{\sqrt{a^2-b^2}}{b}\\
\sqrt{x}&=\frac{a}{b}\\
\sqrt{x+1}&=\frac{\sqrt{a^2+b^2}}{b}
\end{align}
Since, $\sqrt{x+1}$ is rational, then $\sqrt{a^2+b^2}$ must be an integer. So,
$$a^2+b^2=c^2$$
which implies
\begin{align*}
&a=p^2-q^2\\
&b=2pq\\
&c=p^2+q^2
\end{align*}
for some integers $p,q,r$.
Similarly, $\sqrt{a^2-b^2}$ should also be an integer. So,
$$(p^2-q^2)-4p^2q^2=x^2$$
for some integer $x$, which is not possible because of this.
This completes the proof.
Is that okay?
