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 .
26 Aleksander Grytczuk From (15) of Lemma 2 and the assumption that p \ yo we obtain (26) p 2 \P 2 r+P 2 r_ x. By (25), (26) and the fact that p | L,L = cQ r + it follows that P I bQ r — cQ r_i. Now, we can prove the converse of the theorem. Assume that (27) p\cQ r + bQ r-i, p I bQ r — cQ r-i. Prom (15) of Lemma 2 and Lemma 3 we obtain (28) P 2 + P 2_ x = p(Q 2 r + Q 2 T_ X) = pQ s = py 0. By (27) and (25) it follows that p 2 \ P 2 + P 2_ x and therefore from (28) we get p I y 0. The proof of the Theorem 1 is complete. Proof of the Theorem 2. Prom Lemma 3 we have P s_i = P rQ r + Pr-iQr-i • Substituting (10) and (11) of Lemma 2 to this equality we obtain (29) P s_! - b(Q 2 r - Q 2 r_ x ) + 2cQ rQ r. 1 . By (29) easily follows that (30) P 2_,+l = b 2(Q 2 r-Q 2 r_ 1) 2+4bcQ rQ r^ l(Q 2 r-Q 2 r_ l) + 4c 2Q 2 rQ 2 r_ l+l. On the other hand from Lemma 2 we can deduce that (31) c(Q 2 r - Ql + (-l) r+ 1 - 2&Q rQ r_i. Prom (30) and (31) we obtain (32) C 2(^ 2-i+l) = (b 2+c 2) (4(ü 2 + c 2)QIqI_ 1 — 4b(—l) r+ 1 Q rQ r-i + l) . Since (x 0,y 0) = ( P s-i,Q s-i ) then P 2_ t + 1 = pQ\_ x. Suppose that p \ y 0. Then we have (33) P 3\Ph + IBy (33) and (32) it follows that (34) p\4bQ rQ r. l-(-l) r +\
