By Lemma 3, 4 m$_{p}$ ( x ) = q$_{1}$ ( x ) q − $_{1}$( x ) = f ( x ) 2 − ( p ∗ ) g ( x ) 2 And so, replacing x by 1, we get (3.2) 4 p = x 2 1 − p ∗ y 2 1 where x$_{1}$ = f (1), y$_{1}$ = g (1) Since f ( x ) , g ( x ) ∈ Z [ x ], x$_{1}$, y$_{1}$ ∈ Z , and we can see that Lemma 3 relates to Pell’s Equation insofar as it gives us a pair ( x$_{1}$, y$_{1}$ ) that verifies an equation very similar to (3.1). 4 p = x 2 1 − p ∗ y 2 1 ⇒ x 2 1 = 4 p + p ∗ y 2 1 ⇒ p | x 2 1 ⇒ p | x$_{1}$ since p is prime. So letting pξ$_{1}$ = x$_{1}$ , we can rewrite equation (3.2) as 4 p = p $^{2}$ξ 2 1 − p ∗ y 2 $_{1}$, and so, dividing by p , (3.3) pξ 2 1 − ( − 1) $^{p}$− 1 2 y 2 1 = 4 We now analyze q$_{1}$ ( x ) and q − $_{1}$( x ) to obtain some insight as to the values x$_{1}$ and y$_{1}$ . x 2 ≡ ( p − x ) 2 ( p ), so all quadratic residues are in { x 2 ( p ) : 1 ≤ x ≤ $^{p}$− 1 2 } . We can therefore reorder the terms in q$_{1}$ ( x ) and write it as 2 $^{p}$− 1 2 ∏ k =1 ( x − ζ k $^{2}$), and so q$_{1}$ (1) = 2 $^{p}$− 1 2 ∏ k =1 (1 − ζ k $^{2}$). The value of p ∗ depends on the value of p modulo 4 so we will consider the two cases separately for simplicity. ON THE APPLICATIONS OF CYCLOTOMIC FIELDS IN INTRODUCTORY NUMBER THEORY 9 Then (3.3) becomes pξ 2 1 − y 2 1 = 4 (or, to emphasize the similarity to Pell’s Equation, y 2 1 − pξ 2 1 = − 4). Case 1: p ≡ 1 (4). We then have two subcases. If p ≡ 1 (8), then y 2 1 − ξ 2 1 ≡ 4 (8). Trivially y$_{1}$ and ξ$_{1}$ must either be both odd or both even. But 1 2 ≡ 3 2 ≡ 5 2 ≡ 7 2 ≡ 1 (8), so if y$_{1}$ and ξ$_{1}$ were both odd we would have y 2 1 − ξ 2 1 ≡ 0 (8) ⇒ contradiction. It follows that y$_{1}$ and ξ$_{1}$ are both even, and we can thus write y$_{2}$ = y$_{1}$ 2 , ξ$_{2}$ = ξ$_{1}$ 2 ∈ Z . Then y$_{2}$ − pξ 2 2 = − 1. We can use the fact that ( √ p ) 2 ∈ Z to get rid of the minus sign in front of 1. y 2 − pξ 2 2 = − 1 yields ( y$_{2}$ − √ pξ$_{2}$ )( y$_{2}$ + √ pξ$_{2}$ ) = − 1, and so ( y$_{2}$ − √ pξ$_{2}$ ) $^{2}$( y$_{2}$ + √ pξ$_{2}$ ) 2 = 1. But ( y$_{2}$ ± √ pξ$_{2}$ ) 2 = a ± b √ p , a, b ∈ Z . Taking ( x, y ) = ( a, b ),