Proof of $\sqrt{n^2-4}, n\ge 3$ being irrational

Question

Is the proof of $n\ge 3$, $\sqrt{n^2-4} \notin \mathbb{Q} \ \text{correct}$?

$\sqrt{n^2-4} \in \mathbb{Q} \\ \sqrt{n^2-4} = \frac{p}{q} \\ (\sqrt{n^2-4})^2 = \left(\frac{p}{q}\right)^2 \\ q^2\left( n^2-4\right)=p^2 \\ \text{p is divisible by} \left (n^2-4 \right) \Rightarrow p=k\left (n^2-4 \right) \\ q^2 \left (n^2-4 \right)=k^2 \left(n^2-4 \right)^2 \\ q^2=k^2 \left (n^2-4 \right )\Rightarrow \text{it follows that q is also divisible by} \left (n^2-4 \right) \\ \text{Therefore p and q are not co-prime therefore} \Rightarrow \sqrt{n^2-4} \notin \mathbb{Q} \ \square $

Answer 1

You do not give sufficient reason why $p$ is divisible by $n^2-4$ should hold. Consider this modified "proof" of $n\ge 3 \implies \sqrt{n^2-5} \notin \mathbb{Q}$:

$$ \sqrt{n^2-5} \in \mathbb{Q} \\ \sqrt{n^2-5} = \frac{p}{q} \\ (\sqrt{n^2-5})^2 = \left(\frac{p}{q}\right)^2 \\ q^2\left( n^2-5\right)=p^2 \\ \text{p is divisible by} \left (n^2-5 \right) \Rightarrow p=k\left (n^2-5 \right) \\ q^2 \left (n^2-5 \right)=k^2 \left(n^2-5 \right)^2 \\ q^2=k^2 \left (n^2-5 \right )\Rightarrow \text{it follows that q is also divisible by} \left (n^2-5 \right) \\ \text{Therefore p and q are not co-prime therefore} \Rightarrow \sqrt{n^2-5} \notin \mathbb{Q} \ \square$$

This proof must be wrong because $n=3$ leads to $\sqrt{n^2-4}=2\in\mathbb Q$.

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How can the proof be repaired? First show that $n^2-4$ is not a perfect square (which would be a desaster for the claim, as $n^2-4=m^2$ implies $\sqrt{n^2-4}=m\in\mathbb Q$). So assume $n^2-4=m^2$ with $0\le m<n$. Then $4=n^2-m^2=(n+m)(n-m)$. Compare this with the few possible factorizatons of $4$: The possibility $n+m=4$, $n-m=1$ leads to $n=\frac 52\notin\mathbb N$; the posssibility $n+m=n-m=2$ leads to $n=2$, which fails the important given condition $n\ge 3$. So we conclude that $n^2-4$ is not a perfect square. Can you show (or do you already know) the important theorem:

If the positive integer $N$ is not a perfect square, then $\sqrt N\notin\mathbb Q$.

Hint: One can write $N=m^2k$ where $k$ is a product of one or more distic primes. Use one of these primes to show that $\sqrt k$ is irrational.

Answer 2

The fault is when you deduce that $p$ is divisible by $n^2-4$ from the fact that $$ q^2(n^2-4)=p^2 $$ which only implies that $p^2$ is divisible by $n^2-4$.

This would be true if $n^2-4$ is prime, but it isn't in general. For instance, if $n=4$ and $p=6$, $p^2=36$ is divisible by $n^2-4=12$, but $6$ is not divisible by $12$.