Az Eszterházy Károly Tanárképző Főiskola Tudományos Közleményei. 1997. Sectio Mathematicae. (Acta Academiae Paedagogicae Agriensis : Nova series ; Tom. 24)
GRYTCZUK, A., Remark on Ankeny, Artin and Chowla conjecture .
Remark on Ankeny, Artin and Chowla conjecture 27 because p = b 2 + c 2. Now, we can assume that the relation (34) is satisfied. Using (32) we obtain (35) p 2 I c 2(PjU + 1). Since p = b 2 + c 2 and (p, c) = 1, by (35) it follows that (36) p 2 I PU + IBut P 2_i + 1 = pQl-i and consequently from (36) we obtain p \ Q s~\, Qs-i = yo- The proof of the Theorem 2 is complete. Prom Theorem 1 we obtain the following: Corollary. Let (xq , yo) be fundamtental solution of the equation x 2 — py 2 = — 1, where p = 1 (mod 4) is a prime such that p = b 2 + c 2 and let yjp = [i?o; <?i j <72j • • • 5 <?s]> 5 = 2r + 1 be the representation of yfp as a simple continued fraction. If p | yo then ord p(cQ r — 6Q r_i) = 1 or ord p(bQ r — cQ r-1) = 1. Proof. If p I yo then by the Theorem 1 it follows that a = ord p(cQ r + bQ r_I ) > 1 and ß = ord p(bQ r — cQ r_i ) > 1. Suppose that a > 2 and ß > 2. Then we have (37) p 2 I cQ r + 6Q r_i, p 2 \ bQ r - cQ r-i. Prom (37) we obtain p 2 | c 2Q r + bcQ r-i and p 2 \ b 2Q r — bcQ r-\. Hence (38) p 2 I (b 2 +c 2)Q r. Since p = b 2 + c 2 then by (38) it follows that p | Q r. By y 0 = Q s-1 = Ql + Qr-i ^^ virtue of p I yo, p \ Q r we get p \ Q r-i- On the other hand from Lemma 2 we have P T — bQ r + cQ r_\ and therefore we obtain p | P r. Hence we have p | P T and p \ Q r, which is impossible because (P r,Q r ) = 1. The proof is complete. Remark. If the representation of \fd as a simple continued fraction has the period 5 = 3 then d\yo, where (xo,yo) is the fundamental solution of the non-Pellian equation x 2 — dy 2 = —1. Really, putting s = 3 in Lemma 3 we obtain (39) y 0 = Ql + Ql = l + ql
