The equation can be solved using cyclotomy and quadratic residues. A partial solution was found by Dirichlet using this method, building upon the work of Gauss [3]. In this section, we build upon Dirichlet’s work, explicitly writing the solution and using the modern machinery of Galois Theory to streamline the approach. Again, we let p be an odd prime, and define p ∗ = ( − 1) $^{p}$− 1 2 p , i = √ − 1, and start by introducing an important lemma. Lemma 3.      q$_{1}$ ( x ) = 2 ∏ $^{1}$≤ k<p ( k $_{p}$) =1 ( x − ζ $^{k}$) = f ( x ) + √ p ∗ g ( x ) q − $_{1}$( x ) = 2 ∏ $^{1}$≤ k<p ( k $_{p}$) $^{=}$−1 ( x − ζ $^{k}$) = f ( x ) − √$_{p}$$_{∗}$$_{g}$$_{(}$$_{x}$$_{)}$ where f ( x ) , g ( x ) are polynomi- als in Z [ x ] . Proof. Note that the product of the 2 above polynomials (on the left-hand side) is 4 ∏ $^{1}$≤ k<p ( x − ζ $^{k}$) = 4 m$_{p}$ ( x ) ∈ Z [ x ]. It is therefore fixed by any Galois automorphism in Gal ( K : Q ). Now taking θ = ζ$^{$_{p}$2}$ − 1 8 $^{p}$− 1 2 ∏ k =1 (1 − ζ $^{k}$) $^{2}$, we see that θ 2 = p ∗ since ( − 1)$^{$_{p}$2}$ − 1 8 ≡ ( 2 p ) (mod 2), and trivially θ ∈ O $_{K}$. So √$_{p}$$_{∗}$ ∈ O $_{K}$, Now an automorphism σ in the Galois group fixes p ∗ if and only if σ is a square. But this is if and only if σ fixes all (and only) the ζ k such that k is a quadratic residue modulo p . So ∏ $^{1}$≤ k<p ( k $_{p}$) =1 ( x − ζ $^{k}$) ∈ L [ x ] where L = Q $_{(}$√$_{p}$$_{∗}$$_{).}$ All the coefficients in L [ x ] are of the form a + b √ p ∗ where a and b are both rational, and 1 2 · an algebraic integer (allowing for the fact that p ∗ ≡ 1 (4)). The coefficients of 2 ∏ $^{1}$≤ k<p ( k $_{p}$) =1 ( x − ζ $^{k}$) are therefore rational algebraic integers and thus in Z . We can now expand q$_{1}$ ( x ) and rewrite it as q$_{1}$ ( x ) = f ( x $_{) +}$√$_{p}$$_{∗}$$_{g}$$_{(}$$_{x}$$_{) where}$ f ( x ) , g ( x ) are polynomials in Z [ x ]. A similar argument shows that q − $_{1}$( x ) ∈ L [ x ]. Now let τ be the Galois automorphism in Gal ( K : Q ) defined by τ $_{(}$√$_{p}$$_{∗}$$_{) =}$ − √$_{p}$$_{∗}$ (noting that K : L : Q is a tower of fields). Then by the above, and since τ 2 must fix q$_{1}$ ( x ), we must have that τ ( ζ $^{k}$) = ζ l where ( k p )( l p ) = − 1. So since τ is a Galois automorphism over K , we must have τ ( q$_{1}$ ( x )) = q − $_{1}$( x ). This yields that q − $_{1}$( x ) = f ( x ) − √$_{p}$$_{∗}$$_{g}$$_{(}$$_{x}$$_{).}$ □ We will primarily consider the case where d is an odd prime. Pell’s Equation then becomes (3.1) x 2 − py 2 = 1 8 KABALAN GASPARD