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1 k+ 2 k + ... + x k + R(x)=y z Correction to On the Diophantine equation

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Correction to

On the Diophantine equation 1 k + 2 k + ... + x k + R ( x ) = y z

M. VOORHOEVE,

Technische Unioersiteit Eindhoven, Netherlands

b y K. GYORY

University of Debrecen Debrecen, Hungary

and R. TIJDEMAN RUksuniversiteit Leiden

Leiden, Netherlands

(Acta Mathematica, 143 (1979), 1-8)

In the above article the authors claim that a polynomial P with rational integer coefficients which is congruent (mod 4) to

3xS + 2x6 + x4 + 2x 2

has at least three simple roots. Their argumentation is incorrect. In this corrigendum, they wish to repair this defect by proving claim (i) in case B of Lemma 4 in a correct way.

Suppose P can be written as

P(x) - Q(x) T2(x), (*)

Q(x)=-x 2 or Q ( x ) - x 2 + x or Q ( x ) - x 2 + l (rood2),

since x2+x+ 1 does not divide P (mod2). In the first case T(x)--x3+x (mod2), hence T2(x)=-x6+2x4+x 2 (rood4) which does not divide P (mod4). In the second case, the with deg Q~<2.

If d e g Q = 0 , then clearly Q is an odd constant, so T2(x)---xS+x 4 (rood2), hence T(x)=-x4+x 2 (mod2) and TZ(x)=--xS+2x6+x 4 (mod4), which is clearly not the case. If deg Q= 1, then either Q ( x ) - x or Q ( x ) - x + 1 (rood 2). In both cases, the quotient of P and Q cannot be written as a square (mod2). If deg Q=2, then either

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152 M. VOORHOEVE, K. GYORY AND R. TIJDEMAN

quotient of P and Q is not even a square (mod 2). In the third case T(x)-x3+x 2 (mod 2), hence T2(x)=-x6+2xS+x4 (mod4) which does not divide P (mod4). We conclude that P cannot be written in the form (*) with deg Q<3, proving our claim.

Acknowledgements. The authors are indebted to J. Urbanowicz for pointing out the fallacy in the original paper and for showing them a way to repair the defect.

Received February 23, 1987

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