Az Eszterházy Károly Tanárképző Főiskola Tudományos Közleményei. 1998. Sectio Mathematicae. (Acta Academiae Paedagogicae Agriensis : Nova series ; Tom. 25)
GRYTCZUK, A., On a conjecture about the equation Amx + Amy=Amt
64 Aleksander Grytczuk where I = ^ ^ ^ Let {A n } be the recurrence sequence associated with the matrix A. Then applying Lemma 1 to (3) we obtain a (Am(x-Z)2 + ^m(y-z)2) ~ (det A) (A m( x_ z)3 + A m( yz)3) = 1. (6) ^ (Am{x-z)-2 + ^m(y-z)2) = C + ^m{y-z)2) = 0? ii (.4 m( r_ z)_2 + - (det /1) (A m( x_ z)_3 + = 1. Prom Lemma 1, (4) and (5) we obtain similar formulae to (6). Suppose that b / 0 or c ^ 0. Then from (6) we get det A — ±1. On the other hand since A ^ 0, therefore from Lemma 2 we can deduce that (7) /ln-2 = ~^=(a n-ß n). Substituting (7) to (6) we obtain (8) a m{ xz ) + a m( y2 ) = ßMx-z) + pm(y-z) = ^ By (4) and (5) we similarly have ^ a m( z~y ) _ Qj^ix-y) _ ßm(z-y) _ ßm(x-y) _ ^ and (1Q) - — ß m( z~ x) _ ^rn(y-x) _ Erom (8)-(10) it follows that in all cases (11) a rn x + a m y = a m z and ß m x + ß m y = ß m z for natural numbers x,y,z and m > 2, which can be written in the forms (12) + a m( yz ) = 1 and ß + ß m( y-^ = 1. Since A ^ 0, thus we consider two cases: A > 0 or A < 0. Let us suppose that A > 0. Since A — r 2 + 4s and s = - det A - ±1, so we have zl > 5. If r > 0 then we obtain , \ r + y/Ä 1 + Vb r(13) a = —- > ——— > y/2 > 1.
