The diophantine equation \(x^ 2=4q^{a/2}+4q+1\), with an application to coding theory
Let q be a prime power, and x, m positive integers. Using the combination of a method of \textit{F. Beukers} [Acta Arith. 38, 389-410 (1981; Zbl 0371.10014)] with congruence arguments and with a method using second order recurrence sequences it is proved that the only solutions of the equation \(x^ 2=4q^ m+4q+1\) are \(m=2\), \(x=2q+1\); \(q=3\), \((m,x)=(1,5)\) and (3,11) as well as the solutions of the equation \(x^ 2=4q^ m+4q^ 2+1\) with m odd are \(m=1\), \(x=2q+1\) and \(q=2\), \((m,x)=(1,5)\), (3,7) and (7,23). A consequence of the above result is that the only possible parameters of an [n,k] code over the finite field GF(q) with exactly two non-zero weights \(w_ 1\) and \(w_ 2\) and with the property that the minimum weight of the dual code is at least 4 are the following: \[ \begin{alignedat}{5} \text{for any }q&: &\qquad& k=2,&\qquad& n=2,&& w_ 1=1,&& w_ 2=2, \\ & && k=4,&& n=q^ 2+1,&\qquad& w_ 1=(q-1)q,&\qquad& w_ 2=q^ 2; \\ q=2&, && &&n=2^{k-1},&& w_ 1=2^{k-2},&& w_ 2=2^{k-1}; \\ q=3&, && k=5,&& n=11,&& w_ 1=6,&& w_2=9, \\ & && k=6,&& n=56,&& w_ 1=1,&& w_ 2=2; \\ q=4&, && k=6,&& n=78,&& w_ 1=56,&& w_ 2=64, \\ & && k=7,&& n=430,&& w_ 1=320,&& w_ 2=352; \\ q=2^ m&,&& k=3,&& n=2^ m+2,&& w_ 1=2^ m,&& w_ 2=2^ m+2.\end{alignedat} \] The authors confirm so a conjecture of \textit{R. Calderbank} [J. Lond. Math. Soc., II. Ser. 26, 365-384 (1982; Zbl 0545.94014)].
scientific article; zbMATH DE number 3989441
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