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

66 Aleksander Grytczuk .(z-x) follows. It is easy to see that (/?')

m(a

r""

z

) - ([ß')~

2^j 2 . On the other 1 j 1 t ioi\ —2 .It hand we have (ß') < \ and we obtain (ß'yni*-*) = (iß') -2 t(z-x) < <2' because 77

1 > 1. Therefore from (18) we get {ßl

)m(y-z) + x = {ßl

)m{*-z) < which is impossible. In a similar way we obtain a contradiction in the case x — z > 0 and y —

2 < 0. It remains to consider the case when both differences x — z and y — z are negative. Prom (15) we have (19) 1 = < ^ßtyn(y-z) On the other hand we have m(z-x

) / 1 \ " 1 (20) (/}')"(«-•) = (0S'>-*) 2 < (j) <2 and (21) (/3'r'"2» + {(ßr

2)^

1 < Q)-^ < Hence, by (19)-(21), we get a contradiction. Further on we have to consider the case r = 0. But in this case we have a = 1, /5 = —1 and we can can observe that (12) is impossible. Now, we can consider the case A < 0. Since s = — det A = ±1 and A = r

2 + 4s < 0. therefore we have s = — 1 and the inequality r

2 — 4 < 0 implies —2 < r < 2, that is, r = —1,0,1. The case r = 1 is impossble by the assumptions on the eigenvalues of the matrix A. If r = 0 then we obtain that a — i, ß = —i and it is easy to check that (12) does not hold. If r = — 1 then a = ~

1 1 ^ is the third root of unity. Analyzing the exponents m(x — z) and m(y — z) modulo 3 in (12) we get a contradiction.