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

On a conjecture about the equation A

M

X +A

M

Y —A

M

Z 67 Summarizing, we obtain that in the case b / 0 or c / 0 the equation (•) has no solution in positive integers x,y,z and m > 2. So, b = c = 0 and the matrix A can be reduced to a diagonal matrix of the form A = a ® On the other hand for every natural number k we have a 0\

k (a

k 0 0 d (22) A - 1 0 d) ~ V 0 d

k If (•) is satisfied then, by (22), it follfows that (23) a

m

x + a

m

y = a

m

z

, d

m

x + d

m

y = d

m

z

. Prom the assumption of Lemma 3 we have 5 = — det A / 0. This condition implies ad ^ 0, because det A — det ^ ^ ^ = ad. Therefore (23) does not hold. Considering all of the cases the proof of Lemma 3 is complete. Now, we can prove the following. Lemma 4. Let A — ( a ) be an integral matrix and let r = v c d

, Tr A, 5 = - det A and A = r

2 + is. If s f 0 and A = 0, then (*) has no solutions in positive integers x,y,z and m > 2. Proof. Since 6/0, therefore using Lemma 1 in similar way as in the proof of Lemma 3, for the case b / 0 or c / 0 we obtain s = — det A = ±1. Since, A — r

2 + 4s — 0, thus s — — 1 and consequently r

2 — 4 = 0, so we have r = ±2. Therefore we get a = /3=| = lifr = 2 and a = ß = —1 if r — — 2. Prom the well-known theorem of Schur it follows that for any given matrix A there is an unitary matrix P such that (24) A = P*TP, where T is the upper triangular matrix having on the main diagonal the eigenvalues of the matrix A. Suppose that the matrix A = ^ ^ j with integer entries has the eigenvalues a,ß. Prom (24) by easy induction we obtain k r>*T>fc (25) A

K = p*T

KP