Proof verification: finding all prime numbers in the form of $n^3-1, n>1$
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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
edited Dec 18 '18 at 6:36
Eevee Trainer
6,79311237
asked Nov 16 '18 at 14:25
Marko ŠkorićMarko Škorić
70610
$endgroup$
add a comment |
$begingroup$
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
edited Dec 18 '18 at 6:36
Eevee Trainer
6,79311237
asked Nov 16 '18 at 14:25
Marko ŠkorićMarko Škorić
70610
$endgroup$
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
add a comment |
$begingroup$
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
edited Dec 18 '18 at 6:36
Eevee Trainer
6,79311237
asked Nov 16 '18 at 14:25
Marko ŠkorićMarko Škorić
70610
$endgroup$
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
edited Dec 18 '18 at 6:36
Eevee Trainer
6,79311237
asked Nov 16 '18 at 14:25
Marko ŠkorićMarko Škorić
70610
edited Dec 18 '18 at 6:36
Eevee Trainer
6,79311237
asked Nov 16 '18 at 14:25
Marko ŠkorićMarko Škorić
70610
edited Dec 18 '18 at 6:36
Eevee Trainer
6,79311237
edited Dec 18 '18 at 6:36
Eevee Trainer
6,79311237
edited Dec 18 '18 at 6:36
Eevee Trainer
6,79311237
6,79311237
asked Nov 16 '18 at 14:25
Marko ŠkorićMarko Škorić
70610
asked Nov 16 '18 at 14:25
Marko ŠkorićMarko Škorić
70610
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
add a comment |
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
add a comment |
1 Answer
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As the others said in the comments: Yes, your proof is valid.
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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
add a comment |
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1 Answer
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1 Answer
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$begingroup$
As the others said in the comments: Yes, your proof is valid.
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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
add a comment |
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As the others said in the comments: Yes, your proof is valid.
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$begingroup$
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.
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– aleph_two
Dec 18 '18 at 5:14
add a comment |
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As the others said in the comments: Yes, your proof is valid.
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As the others said in the comments: Yes, your proof is valid.
answered Dec 18 '18 at 5:13
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aleph_two
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– aleph_two
Dec 18 '18 at 5:14
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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
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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
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– aleph_two
Dec 18 '18 at 5:14
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3
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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