Az Eszterházy Károly Tanárképző Főiskola Tudományos Közleményei. 1995-1996. Sectio Mathematicae. (Acta Academiae Paedagogicae Agriensis : Nova series ; Tom. 23)

MÁTYÁS F., Two problems related to the Bernoulli numbers

26 Ferenc Mátyás First let us deal with the case a) of the Theorem 2. If k = 2 then by (7), (8), (9) and (3) , v / (2n - 1\ (2n -l\

n P

2(n,c) = n(-i x J^i-^ 0 )Bo c ( ( 2n \ ^ (2n\ ^ (2n\ ^ \ 2 3 -f c c follows. From this, investigating the polinomial equality P2(n,c) = (f(n))

m in the case m > 2, we can see that m = 2 and f(n

) = (n + p) , where = -1 ± and c = 1 ± i2\[2. V Now let us consider the case b) of the Theorem 2. If ft = 3 then by (7), (8), (9), and (3) , , ( (2

n - 1\ /2

n - 1\ „ (2

n - 1\ -K-O- Cr)»-CK n

3 9 + c , 7c + 20 c = • • • = b n

2 n + - . 3 6 12 2 If n) = (f{n))

m and m > 2 then m = 3 and f(n) should have the form f[n) = — + -{/J- But it is easy to verify that such complex numbers c don't exist. If k — 4 then B

3 appears on the right side of (8) and (9). But i?

3 = 0 and so P

3(n,c) = P

4(n,c). Therefore P

4(n,c) = {f(n))

m (m > 2) is also unsolvable. Remark. The statement of the Theorem 2 can also be extended for k > 5 too, but it seems, that there is no polynomial /(n) such that Pk(n

, c) = (/(n))

m where m > 2. References [1] B. BUGGISH, H. HARBORTH and O. P. LOSSER

, Aufgabe 790., Elemente der Mathematik

, Nr 4. (1978), 97-98. [2] P. ADDOR, R. WYSS, Aufgabe 813., Elemente der Mathematik, Nr. 6. (1979), 146-147.