Find all primes of the form n^3-1 for natural number'n'

Dear Student,

Please find below the solution to the asked query:

We know : a³ - b³ = ( a - b ) ( a² + ab + b² ) 

So,

( n³ - 1 ) = ( n³ - 1³ ) = ( n - 1 ) ( n² + n + 1 )

So, We get ( n³ - 1 ) as prime number only if one of ( n - 1 ) and ( n² + n + 1 ) is equal to 1 , So

At n² + n + 1 = 1  ,

n² + n = 0 ,

n ( n + 1 ) = 0  , So

n = 0  and  n = - 1 , Both are not a natural numbers ,

And At n - 1 = 1  ,

n  = 2  and at n  = 2 , That is a natural number we get

( n³ - 1 ) = 2³ - 1 =  8 - 1 = 7  , That is a prime number

So,

We get only one prime number that is in form of n³ - 1 where n is a natural number .


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