Now l ≡ 3 (4), so q$_{1}$ ( i ) q − $_{1}$( i ) = 4 m$_{p}$ ( i ) = 4(1 + i + ... + i $^{l}$) = 4 · ((1 + i − 1 − i ) + (1 + i − 1 − i ) + ... + (1 + i − 1)) = 4 i So f ( i ) $^{2}$− p ∗ g ( i ) 2 = f ( i ) $^{2}$+ pg ( i ) 2 = 4 i , and so y 2 $_{2}$(1+ i ∗ ) $^{2}$+ pξ$_{2}$ (1 − i ∗ ) 2 = 2 y 2 $_{2}$i $^{∗}$− 2 pξ 2 $_{2}$i ∗ = 4 i or, dividing by 2 i ∗ = ± 2 i , y 2 − pξ 2 2 = ± 2 ⇒ ( y$_{2}$ + √ pξ$_{2}$ ) $^{2}$( y$_{2}$ − √ pξ$_{2}$ ) 2 = 4 Now y$_{2}$, ξ$_{2}$ are odd, else y 2 − pξ 2 2 ≡ y 2 $_{2}$+ ξ 2 2 ≡ 0 ̸≡ ± 2 (4). So the coefficients of ( y$_{2}$ + √ pξ$_{2}$ ) 2 = ( y 2 + pξ 2 $_{2}$) + 2 y$_{2}$ξ$_{2}$ √ p are even. We can thus write a = ( y 2 $_{2}$+ pξ 2 $_{2}$) 2 , b = y$_{2}$ξ$_{2}$ ∈ Z and get a 2 − pb 2 = ( y$_{2}$ + √ pξ$_{2}$ ) $^{2}$( y$_{2}$ − √ pξ$_{2}$ ) 2 2 · 2 = 4 4 = 1 This solves the equation, where where we can directly compute f ( i ) and g ( i ) To apply this method to the general case of Pell’s Equation (where d is square-free but not necessarily prime), since d is square-free, it can be written as d = r ∏ k =1 p$_{k}$ where the p$_{k}$ ’s are rational primes. So it suffices to study the case where d = pq for primes p and q and deduce the general case by induction. We will not describe said case in depth here since this paper mainly focuses on prime cyclotomic fields, but we remark that taking Q ( ζ$_{pq}$ ), m$_{pq}$ ( x ) = m$_{p}$ ( x ) m$_{q}$ ( x ) ( x $^{pq}$− 1) / ( $^{x}$− 1) (( x $^{p}$− 1) / ( $^{x}$− $^{1))}$·(( x $^{q}$− 1) / ( $^{x}$−q )) = ( x $^{pq}$− 1)( $^{x}$− 1) ( x $^{p}$− 1)( x $^{q}$− 1) which can be shown to be irreducible by a similar method as the simple proof for showing that $^{p}$− 1 ∑ k =0 x k is the minimal polynomial of ζ$_{p}$ in Z [ x ]. Following the same reasoning as in the case where d = p , we can write 4 m$_{pq}$ ( x ) = f ( x ) 2 ± pqg ( x ) 2 where f ( x ) , g ( x ) ∈ Z . The rest of the problem is solved in a similar fashion as well. Using some interesting approximation methods and quadratic number fields, Ireland & Rosen [5] show that x 2 − dy 2 = 1 has infinitely many solutions for any square-free integer ON THE APPLICATIONS OF CYCLOTOMIC FIELDS IN INTRODUCTORY NUMBER THEORY 13 ( a, b ) = ( i ∗ 4 ( pg ( i ) 2 − f ( i ) $^{2}$) , 1 2 g ( i ) f ( i ) )