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

68 Aleksander Grytczuk for every natural number k, where T

k is the upper triangular matrix with the eigenvalues a

k, ß

k on the main diagonal. If (•) is satisfied then, by (25), it follows that ^26) rpmx rpmy rpmz and from (26) we have (27) a

m* + a

m

y = ß

m

x + ß

m

y = ß

m

z

. Since in our case a — ß = ±1 so we can see that (27) does not hold. Therefore we have b = c — 0 and we get a contradiction as we have got it in the last step of the proof of Lemma 3. So the proof of Lemma 4 is complete. Lemma 5. Let A = f

a ^ ^ be an itegral matrix and let r = Tr A, s = — det A and A = r

2 -f 45. If s — 0 and A ^ 0 then the equation (*) has no solution in positive integers x

, y, z and m > 2. Proof. From the assumptions of Lemma 5 it follows that r / 0 and therefore we can use Lemma 1. Since s — 0 so, by Lemma 1, it follows that If (*) is satisfied then from (28) we obtain (29) r

m

x + r

m

y = r

m

z

. Being r ^ 0, it is easy to see that the equation (29) is impossible in positive integers x,y,z and m > 2. This proves Lemma 5. 3. Proof of the Theorem Suppose that the equation (*) has a solution in postive integers x,y,z and m > 2. Then by Lemma 3, Lemma 4 and Lemma 5 it follows that 5 = det A = 0 and r = Tr A = 0 or the matrix A has an eigenvalue a = 1"

t~2

V

^- the case s = r = 0 we have a — —d and s = — det A = — (ad — be) — — (—d

2 — be) = d

2 -f be = 0 and also putting d — —a we have a

2 -f be = 0. On the other hand we have (on^ A* - (

a bV_fa

2+bc b{a + d)\ _ (a

2+bc br \ [óü

) \c d) ~\c(a + d) d

2 + be J ~ \ cr d

2 + be J '