Proving Primality: When ( p^{28} ) and ( p ) are Primes

In number theory, proving that certain expressions are prime can be a challenging task. This article will guide you through a specific scenario: if ( p^{28} ) and ( p ) are primes, how can we prove that ( p^3 - 4 ) is also a prime?

Understanding the Primes Involved

Let's denote ( p ) as a prime number.

We need to analyze the conditions step by step to understand when ( p^3 - 4 ) could also be prime.

Evaluating ( p )

Case 1: If ( p 2 ):

[ p^2 - 8 2^2 - 8 4 - 8 -4 , text{(not prime)} ]

Case 2: For odd primes ( p ): ( p ) can be any prime greater than 2, such as 3, 5, 7, 11, etc.

Analysis of Odd Primes

For ( p 3 ):

[ p^2 - 8 3^2 - 8 9 - 8 1 , text{(not prime)} ]

For ( p 5 ):

[ p^2 - 8 5^2 - 8 25 - 8 17 , text{(prime)} ]

For ( p 7 ):

[ p^2 - 8 7^2 - 8 49 - 8 41 , text{(not prime)} ]

For ( p 11 ):

[ p^2 - 8 11^2 - 8 121 - 8 113 , text{(not prime)} ]

For ( p 13 ):

[ p^2 - 8 13^2 - 8 169 - 8 161 , text{(not prime)} ]

From these observations, it is clear that the only prime ( p ) which allows both ( p ) and ( p^2 - 8 ) to be prime is ( p 3 ).

Conclusion and Further Analysis

If ( p^2 - 8 ) and ( p ) are primes, then ( p^3 - 4 ) is prime only when ( p 3 ).

Moreover, by observing the transformation ( p^{28} p^{2-1} ), we can further analyze the conditions:

Transformation and Modulo Analysis

Given that ( p^{28} ) is prime, it implies that ( p ) must be an odd prime.

Transformation: ( p^{28} p^{2-1} cdot p^{19} ).

By ( 1 ), it follows that:

[ p - 1 , text{is not a multiple of 3} ,

Which means that ( p ) cannot be congruent to ( 1 mod 3 ) or ( 2 mod 3 ).

Therefore, since ( p ) is an odd prime, we conclude:

[ p equiv 0 mod 3 Rightarrow p 3 ]

Hence, we conclude that:

( p 3 ) ,( p^{28} 17 ) ,( p^3 - 4 3^3 - 4 27 - 4 31 ,text{(prime and a Mersenne prime).} )

This shows that ( p 3 ) is the only prime fulfilling the given conditions.