ON THE APPLICATIONS OF CYCLOTOMIC FIELDS IN INTRODUCTORY NUMBER THEORY 7 Subtracting the latter congruence from the former yields (2.2) ζ − $^{c}$x$_{0}$ + ζ $^{1}$− $^{c}$y$_{0}$ − ζ $^{c}$x$_{0}$ − ζ $^{c}$− $^{1}$y$_{0}$ ≡ 0 ( p ) Now an element of O K = Z [ ζ ] is divisible by p if and only if all of the coefficients as a polynomial in ζ are divisible by p . p ∤ x$_{0}$, y$_{0}$ since p ∤ x$_{0}$y$_{0}$z$_{0}$ , so we must check the cases where one of { c, − c, 1 − c, c − 1 } is congruent to − 1 modulo p or where two of { c, − c, 1 − c, c − 1 } are equal modulo p . These cases can be split as follows: • c ≡ 1 ( p ) (so that 1 − c ≡ c − 1 ( p )). Then p | x$_{0}$ ( ζ − 1 − ζ ) ⇒ p | x$_{0}$ as in the previous case ⇒ contradiction. • c ≡ 0 ( p ) (so that c ≡ − c ( p )). Then p | y$_{0}$ ( ζ − ζ − $^{1}$) = y$_{0}$ ($^{p}$ − 2 ∑ i =2 ζ $^{i}$+ 1) ⇒ p | y$_{0}$ (even if p = 3) ⇒ contradiction. • c ≡ 2 − 1 ( p ) (so that c ≡ 1 − c ( p )). Then p | ( y$_{0}$ − x$_{0}$ ) ζ $^{c}$+ ζ − $^{c}$( x$_{0}$ − y$_{0}$ ). So p | ( x$_{0}$ − y$_{0}$ ). We then rewrite 2.1 as x p 0 + ( − z$_{0}$ ) p = ( − y$_{0}$ ) p (since p is odd). Then with the same argument we will get p | ( x$_{0}$ + z$_{0}$ ). But 2.1 yields x p 0 + y p 0 − z p 0 ≡ 0 ( p ) and so x$_{0}$ + y$_{0}$ − z$_{0}$ ≡ 0 ( p ). This yields 3 x$_{0}$ ≡ 0 ( p ). We suppose for now that p > 3. Then this yields p | x$_{0}$ ⇒ contradiction. • Letting one of { c, − c, 1 − c, c − 1 } be congruent to − 1 modulo p will yield one of the coefficients of the terms of (2.2) as ± ( x$_{0}$ − y$_{0}$ ), giving the same contradiction as in the previous case. We therefore obtain a contradiction in all cases. We have, however, supposed that p > 3 . A general study of the case where p = 3 is done elegantly in [4]. 3. An Approach to Pell’s Equation using cyclotomy Pell’s Equation is x 2 − dy 2 = 1, x, y ∈ Z in x and y , where d ∈ Z $^{+}$. d ≤ 0 trivially yields the single solution (1 , 0), and we can consider d to be square-free, since any square factor of d can be incorporated into y .