Find all prime number in shape $n^3-1,n>1$
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Let some number $p$ is prime number and $p=n^3-1$ so $p=(n-1)(n^2+n+1)$ since $p=1cdot p$ and $n^2+n+1>1$ only $n-1=1$ so $n=2$, and $p=7$, is this ok this is only prime number?
prime-numbers prime-factorization
asked Nov 16 at 14:25
Marko Škorić
69810
add a comment |
up vote
0
down vote
favorite
Let some number $p$ is prime number and $p=n^3-1$ so $p=(n-1)(n^2+n+1)$ since $p=1cdot p$ and $n^2+n+1>1$ only $n-1=1$ so $n=2$, and $p=7$, is this ok this is only prime number?
prime-numbers prime-factorization
asked Nov 16 at 14:25
Marko Škorić
69810
2
Looks good. $quad$.
– lulu
Nov 16 at 14:26
2
Yes, that’s correct.
– KM101
Nov 16 at 14:28
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Let some number $p$ is prime number and $p=n^3-1$ so $p=(n-1)(n^2+n+1)$ since $p=1cdot p$ and $n^2+n+1>1$ only $n-1=1$ so $n=2$, and $p=7$, is this ok this is only prime number?
prime-numbers prime-factorization
asked Nov 16 at 14:25
Marko Škorić
69810
Let some number $p$ is prime number and $p=n^3-1$ so $p=(n-1)(n^2+n+1)$ since $p=1cdot p$ and $n^2+n+1>1$ only $n-1=1$ so $n=2$, and $p=7$, is this ok this is only prime number?
prime-numbers prime-factorization
prime-numbers prime-factorization
asked Nov 16 at 14:25
Marko Škorić
69810
asked Nov 16 at 14:25
Marko Škorić
69810
asked Nov 16 at 14:25
Marko Škorić
69810
asked Nov 16 at 14:25
Marko Škorić
69810
asked Nov 16 at 14:25
Marko Škorić
69810
69810
2
Looks good. $quad$.
– lulu
Nov 16 at 14:26
2
Yes, that’s correct.
– KM101
Nov 16 at 14:28
add a comment |
2
Looks good. $quad$.
– lulu
Nov 16 at 14:26
2
Yes, that’s correct.
– KM101
Nov 16 at 14:28
2
2
Looks good. $quad$.
– lulu
Nov 16 at 14:26
Looks good. $quad$.
– lulu
Nov 16 at 14:26
2
2
Yes, that’s correct.
– KM101
Nov 16 at 14:28
Yes, that’s correct.
– KM101
Nov 16 at 14:28
add a comment |
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2
Looks good. $quad$.
– lulu
Nov 16 at 14:26
2
Yes, that’s correct.
– KM101
Nov 16 at 14:28