Homepage Solution manuals Ivan Niven An Introduction to the Theory of Numbers Exercise 3.2.19* (Biquadratic character of $-1$ modulo $p$)

Exercise 3.2.19* (Biquadratic character of $-1$ modulo $p$)

Show that p is a divisor of numbers of both of the forms m 2 + 1 , n 2 + 2 , if and only if it is a divisor of some number of the form k 4 + 1 .

Answers

Proof. If p = 2 , both propositions are true for m = 1 , n = 0 , k = 1 .

The odd prime p is a divisor of numbers of both of the forms m 2 + 1 , n 2 + 2 if and only if 1 and 2 are quadratic residues modulo p , if and only if

( 1 p ) = 1 , ( 2 p ) = 1 .

This is equivalent to

( 1 ) ( p 1 ) 2 = 1 , ( 1 ) ( p 2 1 ) 8 = 1 ,

that is,

p 1 ( mod 4 ) , p 1 ( mod 8 )  or  p 7 ( mod 8 ) .

Since 7 1 ( mod 8 ) , these conditions are equivalent to

p 1 ( mod 8 ) .

This first part shows that for every odd prime p ,

m , p m 2 + 1  and  n , p n 2 + 1 p 1 ( mod 8 ) .

Moreover, by Theorem 2.37, we may characterize the fourth powers modulo p :

( k , k 4 1 ( mod p ) ( 1 ) ( p 1 ) 4 = 1 2 p 1 4 p 1 ( mod 8 ) .

(biquadratic character of 1 ).

We have proved

k , p k 4 + 1 ( m , p m 2 + 1 )  and  ( n , p n 2 + 2 ) ,

that is, p is a divisor of numbers of both of the forms m 2 + 1 , n 2 + 2 , if and only if it is a divisor of some number of the form k 4 + 1 . □

User profile picture
2024-10-25 08:51
Comments