The only prime of the form n^2-4 is 5?

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The discussion presents a proof that the only prime number of the form n^2-4 is 5. It establishes that for a prime p expressed as n^2-4, the factors n+2 and n-2 must be considered. Since n+2 is always greater than 1 for natural numbers, it concludes that n-2 must equal 1, leading to n being 3. Substituting n back into the equation confirms that p equals 5. The proof is affirmed by multiple participants as correct.
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Homework Statement
Prove the assertion below:
The only prime of the form n^2-4 is 5.
Relevant Equations
None.
Proof: Suppose p is a prime such that p=n^2-4.
Then we have p=n^2-4=(n+2)(n-2).
Note that prime number is a number that has only two factors,
1 and the number itself.
Since n+2>1 for ##\forall n \in \mathbb{N}##,
it follows that n-2=1, and so n=1+2=3.
Thus p=n^2-4=3^2-4
=9-4
=5.
Therefore, the only prime of the form n^2-4 is 5.

Above is my proof for this assertion. Can anyone please verify/review it and see if it's correct?
 
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Same here. Correct.
 
fresh_42 said:
Same here. Correct.
Thank you!
 

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