Proof of $\sqrt{n^2-4}, n\ge 3$ being irrational Is the proof of $n\ge 3$, $\sqrt{n^2-4} \notin \mathbb{Q} \ \text{correct}$?
$\sqrt{n^2-4} \in \mathbb{Q}
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\sqrt{n^2-4} = \frac{p}{q}
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(\sqrt{n^2-4})^2 =  \left(\frac{p}{q}\right)^2  
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q^2\left( n^2-4\right)=p^2
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\text{p is divisible by} \left (n^2-4 \right)
\Rightarrow p=k\left (n^2-4 \right)
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q^2 \left (n^2-4 \right)=k^2 \left(n^2-4 \right)^2
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q^2=k^2 \left (n^2-4 \right )\Rightarrow 
\text{it follows that q is also divisible by} \left (n^2-4 \right)
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\text{Therefore p and q are not co-prime therefore} \Rightarrow  \sqrt{n^2-4} \notin \mathbb{Q} \ 
\square
$

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 $12$.
