a$_{k}$ is defined to be a$_{(}$$_{k}$ mod p ) for all k / ∈ { 0 , ..., p − 1 } ( a$_{p}$ − 1 = 0, trivially). And so $^{p}$− 2 ∑ k =0 ( a$_{p}$ − k − a$_{1}$ ) ζ k = u = ζ − $^{t}$u = $^{p}$− 2 ∑ k =0 ( a$_{k}$$_{+}$$_{t}$ − a$_{(}$$_{p}$ − 1)+ $_{t}$) ζ k by 1 and therefore, since this rep- resentation is unique, we get (1.1) a$_{k}$$_{+}$$_{t}$ − a$_{(}$$_{p}$ − 1)+ t = a$_{p}$ − k − a$_{1}$ for all 0 ≤ k ≤ p − 1 Letting k$_{0}$ be the mod p solution to k + t ≡ p − k ( p ), we get a$_{k}$$_{0}$$_{+}$$_{t}$ = a$_{p}$ − k$_{0}$ and so (1.1) yields a$_{(}$$_{p}$ − 1)+ t = a$_{1}$ . (1.1) then becomes (1.2) a$_{k}$$_{+}$$_{t}$ = a$_{p}$ − k = a − k for all 0 ≤ k ≤ p − 1 Since replacing k by − ( k + t ) in (1.2) leaves the equation invariant, we get $^{p}$− 1 2 pairs of equal terms with distinct indices amongst a$_{0}$, ..., a$_{p}$ − 1 (the ’remaining’ term being a$_{k}$$_{0}$$_{+}$$_{t}$ ). Let b$_{1}$, ..., b $^{p}$− 1 2 be representatives of these distinct pairs, and let b$_{k}$$_{0}$$_{+}$$_{t}$ = a$_{k}$$_{0}$$_{+}$$_{t}$ (we have simply selected and reordered the a$_{i}$ ’s). Now by the proof of 1, there is a unique c modulo p such that ζ $^{c}$u is primary, and this c is the solution to ax ≡ b ( p ) where u ≡ a + bλ ( λ $^{2}$) where λ = (1 − ζ ). Now u = $^{p}$− 2 ∑ k =0 a$_{k}$ζ $^{k}$. Writing, as a polynomial, f ( x ) = $^{p}$− 2 ∑ k =0 a$_{k}$x $^{k}$, we can find a and b by finding the coefficients of 1 and x respectively of f (1 − x ) since ζ = 1 − λ . Making elementary use of the Binomial Theorem, we see that f (1 − x ) = $^{p}$− 2 ∑ k =0 a$_{k}$ (1 − x ) k = $^{p}$− 2 ∑ k =0 a$_{k}$ − $^{p}$− 2 ∑ k =0 ka$_{k}$x + ... (we only need the first two terms). So c is the solution to (1.3) ( $^{p}$− 2 ∑ k =0 a$_{k}$ ) x ≡ − $^{p}$− 2 ∑ k =0 ka$_{k}$ ( p ) Which, since a$_{p}$ − 1 = 0, is equivalent to (1.4) ( $^{p}$− 1 ∑ k =0 a$_{k}$ ) x ≡ − $^{p}$− 1 ∑ k =0 ka$_{k}$ ( p ) Now k$_{0}$ + t ≡ p − k$_{0}$ ( p ) ⇒ k$_{0}$ + t ≡ − ( k$_{0}$ + t ) + t ( p ) ⇒ ( k$_{0}$ + t ) ≡ 2 − $^{1}$t ⇒ b$_{k}$$_{0}$$_{+}$$_{t}$ = a$_{k}$$_{0}$$_{+}$$_{t}$ = a$_{2}$$_{−}$ 1 $_{t}$. Finally, note that for a$_{i}$ = a$_{t}$ − i = b$_{l}$ for 1 ≤ l ≤ $^{p}$− 1 2 by (1.2), ia$_{i}$ + ( t − i ) a$_{t}$ − i = tb$_{l}$ . 4 KABALAN GASPARD