Proving the Compositeness of ( p^{q-1} - q 1 ) Using Fermats Little Theorem

Introduction

Proving the compositeness of a number is a fundamental task in number theory, often accomplished through the application of theorems such as Fermat's Little Theorem (FLT). This article explores how to prove the compositeness of the expression ( p^{q-1} - q 1 ) when ( q ) is a prime number and ( p ) is any integer relatively prime to ( q ).

Context and Theorem Background

As a reminder, Fermat's Little Theorem (FLT) states that if ( p ) is a prime number and ( a ) is an integer not divisible by ( p ), then ( a^{p-1} equiv 1 pmod{p} ).

Given this, we explore a more generalized form using the expression ( p^{q-1} - q 1 ).

Proving the Compositeness

Let's delve into the proof step by step:

Step 1: Assume ( q ) is a Prime Number and ( p ) is Relatively Prime to ( q )

Assume ( q ) is a prime number and ( p ) is an integer such that ( gcd(p, q) 1 ). That means ( p ) and ( q ) are relatively prime.

Step 2: Apply Fermat's Little Theorem

According to Fermat's Little Theorem:

( p^{q-1} equiv 1 pmod{q} )

This implies:

( p^{q-1} 1 kq )

for some integer ( k ).

Step 3: Transform the Expression

Now consider the expression ( p^{q-1} - q 1 ):

( p^{q-1} - q 1 )

Substitute ( p^{q-1} ) from the result of FLT:

( (1 kq) - q 1 kq )

This simplifies to:

( p^{q-1} - q 1 kq )

where ( k ) is a non-zero integer. Since ( kq ) is a product of ( k ) and ( q ), it is composite unless ( k 1 ) or ( k -1 ). However, in both cases, ( kq ) remains composite because ( q ) is a prime and ( k ) is an integer 1.

Step 4: Conclusion

Hence, ( p^{q-1} - q 1 ) is never prime, making it a composite number.

Generalization and Proof Expansion

The initial problem provided a similar proof but used a simpler expression ( p^2 - 2 ) to show that ( p^2 - 2 ) is composite when ( p geq 5 ) is a prime number. Using the same logic, we generalized this to ( p^{q-1} - q 1 ) to demonstrate a broader application of the theorem.

Generalization Example

If ( p geq 5 ) is a prime number, we can substitute ( q - 1 ) for 2, yielding:

( p^{q-1} - q 1 )

By applying FLT again:

( p^{q-1} equiv 1 pmod{q} )

This implies:

( p^{q-1} - q 1 equiv 1 - q 1 equiv 2 - q pmod{q} )

Since ( 2 - q ) is non-zero modulo ( q ), ( p^{q-1} - q 1 ) is not a prime as it is always greater than 1 and divisible by ( q ).

Conclusion

Thus, through the application of Fermat's Little Theorem, we have proven the compositeness of ( p^{q-1} - q 1 ) for any integer ( p ) that is relatively prime to ( q ). This method not only strengthens our understanding of prime numbers and compositeness but also showcases the versatility of the theorem in solving related problems.