12 Replacing x by i = √ − 1, we see that x $^{k}$( x $^{l}$− 2 k − 1) depends on whether p ≡ 3 or 7 (8). g ( x ) = 1 $_{2}$√$_{p}$$_{∗}$ ( q$_{1}$ ( x ) − q − $_{1}$( x )) = 1 √$_{p}$$_{∗}$    ∏ $^{1}$≤ k<p ( k $_{p}$) =1 ( x − ζ $^{k}$) − ∏ $^{1}$≤ k<p ( k $_{p}$) $^{=}$−1 ( x − ζ k $_{)}$   And so g ( ζ ) = − 1 √$_{p}$$_{∗}$    ∏ $^{1}$≤ k<p ( k $_{p}$) =1 ( ζ − ζ − k $_{)}$   and g ( ζ − $^{1}$) = 1 √$_{p}$$_{∗}$    ∏ $^{1}$≤ k<p ( k $_{p}$) =1 ( ζ − 1 − ζ k $_{)}$   A similar line of reasoning as for f ( x ) gives us that g ( ζ ) = + ζ $^{l}$g ( ζ − $^{1}$). Following the same steps as for f ( x ), we find that, writing g ( x ) as 1 √$_{p}$$_{∗}$ l ∑ k =0 a$_{k}$x $^{k}$, we get a$_{l}$ − k = + a$_{l}$ for all 0 ≤ k ≤ l (with a$_{l}$ = a$_{0}$ = 0 this time). We can therefore similarly rewrite g ( x ) as $^{l}$− 1 2 ∑ k =0 b$_{k}$x $^{k}$( x $^{l}$− 2 k + 1), b$_{k}$ ∈ Z (remembering that g ( x ) ∈ Z [ x ] by 3). A similar argument shows that g ( i ) = $^{l}$− 1 2 ∑ k $_{=0}$± b$_{k}$ (1 − i ∗ ) = ξ$_{2}$ (1 − i ∗ ) where ξ$_{2}$ ∈ Z . Now, Let p ≡ 3 (8). Then l ≡ 1 (4) and simple calculation yields i $^{k}$( i $^{l}$− 2 k − 1) =    1 − i if k ≡ 1 , 2 (4) − (1 − i ) if k ≡ 0 , 3 (4) p ≡ 7 (8) ⇒ l ≡ 3 (4), and the same type of calculation yields i $^{k}$( i $^{l}$− 2 k − 1) =    1 + i if k ≡ 3 , 2 (4) − (1 + i ) if k ≡ 0 , 1 (4) Writing i ∗ =    − i if p ≡ 3 (8) + i if p ≡ 7 (8) , we see that f ( i ) = $^{l}$− 1 2 ∑ k $_{=0}$± b$_{k}$ (1 + i ∗ ) = y$_{2}$ (1 + i ∗ ) where y$_{2}$ ∈ Z . KABALAN GASPARD