Find all natural numbers $n$ such that $\sqrt{n-1}+\sqrt{n+1}$ is rational...

Question

There aren't any natural number $n$ for which the given condition is satisfied. Here is how I proved it:

For $\sqrt{n-1}+\sqrt{n+1}$ to be rational, $\sqrt{n-1}$ and $\sqrt{n+1}$ must individually be rational as the sum of two positive numbers can't be rational if at least one of them is irrational.

For $\sqrt{n-1}$ and $\sqrt{n+1}$ to be individually rational, $(n-1)$ and $(n+1)$ must be perfect squares. We can see that the two numbers differ by $2$: $(n+1)-(n-1)=2$. This is not possible as consecutive squares differ by at least $3$. $\blacksquare$

The book I am using proves the same conditions in a more involved way:

If $\sqrt{n-1}+\sqrt{n+1}$ is rational, $\frac{1}{\sqrt{n-1}+\sqrt{n+1}}$ must also be. Multiplying by $\frac{\sqrt{n-1}-\sqrt{n+1}}{\sqrt{n-1}-\sqrt{n+1}}$, we get $\frac{\sqrt{n-1}-\sqrt{n+1}}{2}$ is rational $\implies \sqrt{n-1}-\sqrt{n+1}$ is rational.

The proof then proceeds to do what I did but for both $\sqrt{n-1}-\sqrt{n+1}$ and $\sqrt{n-1}+\sqrt{n+1}$.

There is also an answer on Mathematics that proves the same in a way similar to the book prove I just mentioned.

Is the way I have proved the same incomplete or wrong?

My question is not duplicate because it is asking why my proof is wrong. The linked duplicate asks for a proof; I have provided the proof, both mine and the one found in book, with the question.

Answer 1

The sum of two irrational numbers can be rational (e.g. $x,\,q-x$ with $q\in\Bbb Q\not\owns x$). Their strategy was to note that, since $\sqrt{n-1}\pm\sqrt{n+1}\in\Bbb Q$, taking linear combinations thereof gives $\sqrt{n\pm 1}\in\Bbb Q$.

Answer 2

If $\sqrt{n-1}+\sqrt{n+1}$ is rational and $n>1$ then $(\sqrt{n-1}+\sqrt{n+1})^2=2n+2\sqrt{n^2-1}$ is rational. Then $\sqrt{n^2-1}$ is rational. Then $n^2-1=\frac{p^2}{q^2}$, where $p,q \in \mathbb{N}$, $gcd(p,q)=1$. Then $q=1$. Then $n^2-1=p^2$. But $n^2-1=p^2$ hasn't solutions.