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

24 Ferenc Mátyás solved in [1] and the author of this paper also was among the solvers. In our Theorem 1., using the Bernoulli numbers, we give a new proof for this problem. Theorem 1. If m > 2 is an integer then there exists a polynomial fk{n) such that Sk{n

) = for every positive integer n if and only if m = 2, k = 3 and /3(n) = \n

2 + \n. Proof. It is known that Sk{n) can be expressed by the Bernoulli numbers and the binomial coefficients, that is (2) '

x • + „ 2 . fk + 1 + •••+1 f

c_JW + ( k \B

kn moreover (3

) B

0 = l,B

l = l-,B

2 = i,J?

3 = 0,... and Bj — 0 if and only if j > 3 and j is odd. Let fk(n) = CLjU

3 + • • - + 0171 + 00 be a polynomial of n over the rationals and aj í 0. If Sk{n) = ^fk(n)) f°

r some m, then by (2) öq = 0 follows and since m > 2 so the degree of the polynomial (Vfc(n)j is at least two, that is Bk = 0 in (2). But by (3) Bk — 0 implies that k > 3, k is an odd integer and Bk-i / 0. From the equality Sfc(n) = it follows that ÍTI (C10 + • • •

+ (Í - 1) = a>

m'

+ " •

+ and from this we get m = 2 and a\ ^ 0. So we have to investigate the equality w rhr ("

k+1 +'''

+ C -!)

B

k~

in

2)

= {ajn3 +"'

+ain)

2 from which we obtain that k + 1 = 2j and j- = a

2. Moreover dj is a rational number, therefore k + 1 = 4/ and j = 2/. (4) can be written in the following form: 1 (MI + (5) = (a

2jn

2

f + b a

2n

2 + <nn)