Proof verification: finding all prime numbers in the form of $n^3-1, n>1$












0












$begingroup$


Let $p$ be a prime number of the form $p = n ^3 - 1$ for a positive integer $n geq 2$.



Then, factoring the difference of perfect cubes, we obtain $p = (n-1)(n^2 + n + 1)$.



Since $p = 1 cdot p$ as well, and $n^2 + n + 1 > 1$, $n$ must satisfy $n-1=1$, thus implying $n=2$, yielding $p=7$.



Is this proof valid? Is $p=7$ the only prime number of this form?






proof-verification prime-numbers prime-factorization







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  • 3




    $begingroup$
    Looks good. $quad$.
    $endgroup$
    – lulu
    Nov 16 '18 at 14:26






  • 3




    $begingroup$
    Yes, that’s correct.
    $endgroup$
    – KM101
    Nov 16 '18 at 14:28
















0












$begingroup$


Let $p$ be a prime number of the form $p = n ^3 - 1$ for a positive integer $n geq 2$.



Then, factoring the difference of perfect cubes, we obtain $p = (n-1)(n^2 + n + 1)$.



Since $p = 1 cdot p$ as well, and $n^2 + n + 1 > 1$, $n$ must satisfy $n-1=1$, thus implying $n=2$, yielding $p=7$.



Is this proof valid? Is $p=7$ the only prime number of this form?






proof-verification prime-numbers prime-factorization







share|cite|improve this question











$endgroup$








  • 3




    $begingroup$
    Looks good. $quad$.
    $endgroup$
    – lulu
    Nov 16 '18 at 14:26






  • 3




    $begingroup$
    Yes, that’s correct.
    $endgroup$
    – KM101
    Nov 16 '18 at 14:28














0












0








0





$begingroup$


Let $p$ be a prime number of the form $p = n ^3 - 1$ for a positive integer $n geq 2$.



Then, factoring the difference of perfect cubes, we obtain $p = (n-1)(n^2 + n + 1)$.



Since $p = 1 cdot p$ as well, and $n^2 + n + 1 > 1$, $n$ must satisfy $n-1=1$, thus implying $n=2$, yielding $p=7$.



Is this proof valid? Is $p=7$ the only prime number of this form?






proof-verification prime-numbers prime-factorization







share|cite|improve this question











$endgroup$




Let $p$ be a prime number of the form $p = n ^3 - 1$ for a positive integer $n geq 2$.



Then, factoring the difference of perfect cubes, we obtain $p = (n-1)(n^2 + n + 1)$.



Since $p = 1 cdot p$ as well, and $n^2 + n + 1 > 1$, $n$ must satisfy $n-1=1$, thus implying $n=2$, yielding $p=7$.



Is this proof valid? Is $p=7$ the only prime number of this form?






proof-verification prime-numbers prime-factorization




proof-verification prime-numbers prime-factorization






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 18 '18 at 6:36









Eevee Trainer

6,79311237




6,79311237










asked Nov 16 '18 at 14:25









Marko ŠkorićMarko Škorić

70610




70610








  • 3




    $begingroup$
    Looks good. $quad$.
    $endgroup$
    – lulu
    Nov 16 '18 at 14:26






  • 3




    $begingroup$
    Yes, that’s correct.
    $endgroup$
    – KM101
    Nov 16 '18 at 14:28














  • 3




    $begingroup$
    Looks good. $quad$.
    $endgroup$
    – lulu
    Nov 16 '18 at 14:26






  • 3




    $begingroup$
    Yes, that’s correct.
    $endgroup$
    – KM101
    Nov 16 '18 at 14:28








3




3




$begingroup$
Looks good. $quad$.
$endgroup$
– lulu
Nov 16 '18 at 14:26




$begingroup$
Looks good. $quad$.
$endgroup$
– lulu
Nov 16 '18 at 14:26




3




3




$begingroup$
Yes, that’s correct.
$endgroup$
– KM101
Nov 16 '18 at 14:28




$begingroup$
Yes, that’s correct.
$endgroup$
– KM101
Nov 16 '18 at 14:28










1 Answer
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2












$begingroup$

As the others said in the comments: Yes, your proof is valid.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    This answer exists to remove this question from the Unanswered queue. To complete this process, please upvote this answer to a score of at least +1.
    $endgroup$
    – aleph_two
    Dec 18 '18 at 5:14











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1 Answer
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active

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1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









2












$begingroup$

As the others said in the comments: Yes, your proof is valid.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    This answer exists to remove this question from the Unanswered queue. To complete this process, please upvote this answer to a score of at least +1.
    $endgroup$
    – aleph_two
    Dec 18 '18 at 5:14
















2












$begingroup$

As the others said in the comments: Yes, your proof is valid.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    This answer exists to remove this question from the Unanswered queue. To complete this process, please upvote this answer to a score of at least +1.
    $endgroup$
    – aleph_two
    Dec 18 '18 at 5:14














2












2








2





$begingroup$

As the others said in the comments: Yes, your proof is valid.






share|cite|improve this answer











$endgroup$



As the others said in the comments: Yes, your proof is valid.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








answered Dec 18 '18 at 5:13


























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aleph_two













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    This answer exists to remove this question from the Unanswered queue. To complete this process, please upvote this answer to a score of at least +1.
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    – aleph_two
    Dec 18 '18 at 5:14


















  • $begingroup$
    This answer exists to remove this question from the Unanswered queue. To complete this process, please upvote this answer to a score of at least +1.
    $endgroup$
    – aleph_two
    Dec 18 '18 at 5:14
















$begingroup$
This answer exists to remove this question from the Unanswered queue. To complete this process, please upvote this answer to a score of at least +1.
$endgroup$
– aleph_two
Dec 18 '18 at 5:14




$begingroup$
This answer exists to remove this question from the Unanswered queue. To complete this process, please upvote this answer to a score of at least +1.
$endgroup$
– aleph_two
Dec 18 '18 at 5:14


















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