10 KABALAN GASPARD we have solved (3.1). Summarizing, we get a solution from ( a, b ) = ( 1 4 ( g (1) 2 + f (1) 2 p ) , f (1) g (1) 2 p ) where we can directly compute f (1) and g (1) If p ≡ 5 (8), y 2 1 + 3 ξ 2 1 ≡ 4 (8). Given that the only quadratic residues modulo 8 are 0 , 1 , 4, we must have ( y 2 $_{1}$, ξ 2 $_{1}$) ≡ (1 , 1) , (0 , 4) or (4 , 0) (8). We now use the fact that 8 2 = 2 $^{2}$·3 = 4 3 and consider ( y$_{1}$ + √ pξ$_{1}$ ) 3 = ( y 3 1 + 3 pξ 2 $_{1}$y$_{1}$ ) + √ p ( pξ 3 1 + 3 y 2 $_{1}$ξ$_{1}$ ) = y$_{2}$ + √ pξ$_{2}$ and see that y 2 − pξ 2 2 = ( y 2 1 − pξ 2 $_{1}$) 3 = − 4 $^{3}$. But y$_{2}$ = y$_{1}$ ( y 2 $_{1}$+3 pξ 2 $_{1}$) ≡ y$_{1}$ ( y 2 1 − ξ 2 $_{1}$) (8). ( y 2 $_{1}$, ξ 2 $_{1}$) ≡ (1 , 1) (8) ⇒ y$_{2}$ ≡ 0 (8). ( y 2 $_{1}$, ξ 2 $_{1}$) ≡ (0 , 4) or (4 , 0) (8) ⇒ y$_{2}$ ≡ 4 · 4 , 0 · 4 or ± 2 · 4 ≡ 0 (8). So in any case y$_{2}$ ≡ 0 (8). Similarly ξ$_{2}$ = ξ$_{1}$ ( pξ 2 $_{1}$+ 3 y 2 $_{1}$) ≡ ξ$_{1}$ (5 ξ 2 $_{1}$+ 3 y 2 $_{1}$) (8). ( y 2 $_{1}$, ξ 2 $_{1}$) ≡ (1 , 1) (8) ⇒ ξ$_{2}$ ≡ ξ$_{2}$ (5 + 3) ≡ 0 (8). ( y 2 $_{1}$, ξ 2 $_{1}$) ≡ (0 , 4) or (4 , 0) (8) ⇒ ξ$_{2}$ ≡ ± 2 · 4 , 0 · 4 or 4 · 0 ≡ 0 (8). So in any case ξ$_{2}$ ≡ 0 (8). So 8 | y$_{2}$, ξ$_{2}$ and thus, writing y$_{3}$ = y$_{2}$ 8 , ξ$_{3}$ = ξ$_{2}$ 8 ∈ Z , we get ( y 2 3 − pξ 2 $_{3}$) = − 4 3 8 2 = − 1. As in the case where p ≡ 1 (8), writing ( y$_{3}$ ± √ pξ$_{3}$ ) 2 = a ± b √ p , a, b ∈ Z , ( x, y ) = ( a, b ) is a solution of (3.1). Summarizing, we get a solution from ( a, b ) =   1 $_{64}$(( g (1) 3 + 3 f (1) $^{2}$g (1) p ) 2 + p ( f (1) 3 p 2 + 3 g (1) $^{2}$f (1) p ) $^{2}$) , 1 $_{32}$( g (1) 3 + 3 f (1) $^{2}$g (1) p )( f (1) 3 p 2 + 3 g (1) $^{2}$f (1) p )   Case 2: p ≡ 3 (4). Let l = $^{p}$− 1 2 . p ≡ 3 (4) ⇒ l is odd. We see that f ( x ) = 1 $_{2}$( q$_{1}$ ( x ) + q − $_{1}$( x )) = ∏ $^{1}$≤ k<p ( k $_{p}$) =1 ( x − ζ $^{k}$) + ∏ $^{1}$≤ k<p ( k $_{p}$) $^{=}$−1 ( x − ζ $^{k}$). f is of degree l . We shall find a relation amongst the coefficients of f by comparing f ( ζ ) and f ( ζ ) = f ( ζ ) (since f ( x ) ∈ Z [ x ]). Trivially ( 1 p ) = 1, so ∏ $^{1}$≤ k<p ( k $_{p}$) =1 ( ζ − ζ $^{k}$) = 0 and so f ( ζ ) = ∏ $^{1}$≤ k<p ( k $_{p}$) $^{=}$−1 ( ζ − ζ $^{k}$). Also note that ( − 1 p ) = ( − 1) $^{p}$− 1 2 = − 1, and so ( k p ) = − ( − k p ) for all 1 ≤ k ≤ p − 1. So f ( ζ ) = ∏ $^{1}$≤ k<p ( k $_{p}$) =1 ( ζ − ζ − $^{k}$). By the same line of reasoning, f ( ζ ) = f ( ζ − $^{1}$) =