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 conjecturc about the equation A

M

X + A

M

Y = A

,

N

* 65 Prom (13) and (12) it follows that both exponents m(x — z

) and m(y — z) must be negative. On the other hand fom (13) we have a~

2 < \ and by (12) it follows that it cannot happen that both exponents m(x — z) and m(y — z) are < —2. Therefore one of them must be equal to -1 and we obtain m(x — z) = —1 or m(y — z) — — 1. But this is impossible, because m > 2 and x,y,z are positive integers. After this we consider the case r < 0. Let us suppose that r < 0 and put r = —r', where r' > 0. Then we have _ r-y/

A _ _ r' + y/

Ä _ _ 2 2 and . [Ä 1 + y/5 rß = r' + J-> —

y- > V2> 1. Substituting ß = —ß to the second equation of (12) we obtain (14) (-1 (ß')

m<<

xz) + (_!)m(y-z) ^ß^m(y-z) = ^ K rn is even then as in our previous case we obtain a contradiction. So, we can assume that m is an odd natural number greater than 2. If x — z and y — z are odd then it is easy to see that (14) does not hold. Therefore one of them must be even and from (14) we obtain [lb) (ß')

m{

x~

z

) -{ß')

m{

y~

z

] = I, if 2-2 is even and y-z is odd and (16

) ^m(y-z) = ^ y_

z j

s eye

n ^ x _ z is od

d. Because of the symmetry, it is sufficient to consider one of these equations. Let us suppose that (15) is satisfied. If x — z > 0 and y — z > 0 then, by(15), it follows that x — z > y — z. On the other hand, (15) can be represented in the form m(y — z) (/nf\m(x — z) (17) (0/jmiv-*; j^ß^x-z, _ ij = 1. The condition x — z > y—z implies x > y and since ß' > A/2, m>2

1x — Z>0 and y — z > 0, therefore (17) is impossible. Hence we get that one of the differences x — z so y - z must be negative. Suppose that x — z < 0 and y - z > 0. Then from (15) (18) (ß')

mi

xz

) = 4- 1