Proving Prime Factors of the Form 4n 1 for Even ( x^{21} 1 )

Introduction

In number theory, the concept of prime factors and their forms plays a crucial role. Specifically, when analyzing the prime factorization of ( x^{21} 1 ), we can explore under what conditions such a factorization can have prime factors of the form ( 4n 1 ).

The Proposition and Assumptions

Consider the proposition: If a prime ( p ) is of the form ( 4n - 1 ), it does not divide ( x^{21} 1 ) for any integer ( x ).

To prove this proposition, we will use a method of contradiction. Assume the opposite, that there exists an integer ( x ) such that a prime ( p ) of the form ( 4n - 1 ) divides ( x^{21} 1 ).

The Algebraic Proof

Let us assume ( x^{21} 1 equiv 0 pmod{p} ). This implies:

( x^{21} equiv -1 pmod{p} )

Raising both sides to the power of an odd integer ( 2n - 1 ), we get:

( x^{4n-2} equiv -1 pmod{p} )

Continuing this process, we reach:

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

Multiplying both sides by ( x ) gives:

( x^p equiv -x pmod{p} )

By Fermat's Little Theorem, we know:

( x^p equiv x pmod{p} )

Subtracting the congruence, we obtain:

( 0 equiv 2x pmod{p} )

Since ( p ) is an odd prime, this is a contradiction because ( x ) is not divisible by ( p ).

This contradiction establishes that a prime ( p ) of the form ( 4n - 1 ) cannot divide ( x^{21} 1 ).

Quadratic Reciprocity and Conclusion

The proposition we have proven is a step towards understanding the solvability of the congruence ( x^2 equiv -1 pmod{p} ), which is solvable for a prime ( p ) if and only if ( p equiv 1 pmod{4} ).

For an even ( x ), ( x^{21} 1 ) must have prime factors of the form ( 4n 1 ). This is because any prime ( q ) dividing ( x^{21} 1 ) must be of the form ( 4n 1 ), as we have shown that ( q ) of the form ( 4n - 1 ) cannot divide ( x^{21} 1 ).

Nonetheless, for ( x 2 ), ( x^{21} 1 5 ), and for ( x 4 ), ( x^{21} 1 17 ), showing that there can be multiple prime factors of the form ( 4n 1 ).

Thus, the final conclusion is that for an even ( x ) with ( x eq 0 ), all prime divisors of ( x^{21} 1 ) must be of the form ( 4n 1 ).

It is important to note that the result does not hold if ( x 0 ).

Conclusion

We have shown that for an even integer ( x ) and ( x eq 0 ), the number ( x^{21} 1 ) must have prime factors of the form ( 4n 1 ).

Keywords: prime factor, form 4n 1, quadratic reciprocity