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

Ori a conjecture about the equation A

M

I +A

M

Y = A

M

Z 63 Lemma 1. Let A = ( ^ 1 be an integral matrix such that Tr A ^ 0 2. Basic Lemmas b d or det A ^ 0 and let r = a + d = TV A, s = - det A, A

0 = r, Ai = rA

0 + 5 and A

n = rA

n_ 1 + sA

n_

2 if n > 2. Then for every natural number n > 2, we have A

n = a b \ _ f a/

in-2 + sA

n_

3 6i4

n_

2 C Ű? Y ~ I cA

n_

2 dA

n_

2 -F <SA

n_3 where we put A_i = 1. The proof of this Lemma immediately follows from Theorem 1 of [6]. Lemma 2. Let A be an integral matrix satisfying the assumptions of Lemma 1 and let A

n be the recurrence sequence associated with the matrix A as in Lemma 1. Moreover, let A

n be the discriminant of the characteristic polynomial of A

n if n > 2 and let A\ = A — r

2 +45. Then for every natural number n > 2 we have A

n = AA

2

n_

2. The proof of Lemma 2 is given in [4]. Lemma 3. Let A = ^^ ^ j be an integral matrix and let f(x) = x

2 — (Tr A)x + det A be the characteristic polynomial of A with the roots a, ß ^ and the discriminant A = r

2 + 4s, where r = a + d= TrA and s = - det A. If s ^ 0 and A ^ 0 then the equation (•) has no solutions in natural numbers x,y,z and m > 2. Proof. If x = z and (•) is satisfied then A

m

y = 0, thus det A = 0, which contradicts to our assumption. Similarly we obtain a contradiction when y = z. If x = y then by (•) it follows that 2A

m

r = A

771

2, hence 4(det A)

m

x — (det A)

m

2 and so we obtain a contradiction, because the last equality is impossible in natural numbers x,y,z and m > 2 with integer det A ± 0. Further on we can assume that if (*) is satisfied, then x,y and z are distinct natural numbers. Since s = — det A ^ 0. therefore there exists the inverse matrix A1 and from (•) we obtain (3) Am(x-z) + Am(y-z) = mi

n{

a.

j = ^ (4) Am(x-y) +I= Am(z-y)^ min{x

, y

, z} = V

, (5) 1 + A

m(

y~

x

) = A

m(

z"

x

), if min{x,y,z} = x,