The only prime of the form n^3-1 is 7?

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The only prime number of the form n^3-1 is definitively 7. The proof establishes that if p is a prime such that p=n^3-1, then it can be factored as (n-1)(n^2+n+1). Since n^2+n+1 is greater than 1 for all natural numbers n, the only solution occurs when n-1 equals 1, leading to n=2 and p=7. This conclusion is validated by the mathematical reasoning presented in the discussion.

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Homework Statement
Prove the assertion below:
The only prime of the form n^3-1 is 7.
[Hint: Write n^3-1 as (n-1)(n^2+n+1).]
Relevant Equations
None.
Proof: Suppose p is a prime such that p=n^3-1.
Then we have p=n^3-1=(n-1)(n^2+n+1).
Note that prime number is a number that has only two factors,
1 and the number itself.
Since n^2+n+1>1 for ##\forall n\in\mathbb{N}##,
it follows that n-1=1, and so n=1+1=2.
Thus n=2, and so p=n^3-1=2^3-1=8-1=7.
Therefore, the only prime of the form n^3-1 is 7.

Above is my proof for this assertion. Can anyone please verify/review it and see if it's correct?
 
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Math100 said:
Homework Statement:: Prove the assertion below:
The only prime of the form n^3-1 is 7.
[Hint: Write n^3-1 as (n-1)(n^2+n+1).]
Relevant Equations:: None.

Above is my proof for this assertion. Can anyone please verify/review it and see if it's correct?
It is correct, and your wording is improving!
 
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fresh_42 said:
It is correct, and your wording is improving!
Thank you!
 

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