Page:On the expression of a number in the form π‘Žπ‘₯Β²+𝑏𝑦²+𝑐𝑧²+𝑑𝑒².djvu/7

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in the form \scriptstyle{ax^2+by^2+cz^2+du^2}
  1. (6Β·4) \scriptstyle{n\equiv 3\pmod{4}}.

    There is only a finite number of exceptions. Take

    \scriptstyle{N=4^\lambda(8\mu+7)}.


    If \scriptstyle{\lambda\geq 1}, take \scriptstyle{u=1}. Then

    \scriptstyle{N-nu^2\equiv 1\ \mathit{or}\ 5\pmod{8}}.


    If \scriptstyle{\lambda=0}, take \scriptstyle{u=2}. Then

    \scriptstyle{N-nu^2\equiv 3\pmod{8}}.


    In either case the proof is completed as before.

    In order to determine precisely which are the exceptional numbers, we must consider more particularly the numbers between \scriptstyle{n} and \scriptstyle{4n} for which \scriptstyle{\lambda=0}. For these \scriptstyle{u} must be \scriptstyle{1}, and

    \scriptstyle{N-nu^2\equiv 0\pmod{4}}.


    But the numbers which are multiples of \scriptstyle{4} and which cannot be expressed in the form \scriptstyle{x^2+y^2+z^2} are the numbers

    \scriptstyle{4^\kappa(8\nu+7),\quad(\kappa=1,~2,~3,~\ldots,\,\nu=0,~1,~2,~3,~\ldots)}.


    The exceptions required are therefore those of the numbers

    \scriptstyle{n+4^\kappa(8\nu+7)} (6Β·41)


    which lie between \scriptstyle{n} and \scriptstyle{4n} and are of the form

    \scriptstyle{8\mu+7} (6Β·42).


    Now in order that (6Β·41) may be of the form (6Β·42), \scriptstyle{\kappa} must be \scriptstyle{1} if \scriptstyle{n} is of the form \scriptstyle{8k+3} and \scriptstyle{\kappa} may have any of the values \scriptstyle{2,~3,~4,~\ldots} if \scriptstyle{n} is of the form \scriptstyle{8k+7}. Thus the only numbers which cannot be expressed in the form (5Β·2), in this case, are those of the form \scriptstyle{4^\lambda(8\mu+7)} less than \scriptstyle{n} and those of the form

    \scriptstyle{n+4^\kappa(8\nu+7),\quad(\nu=0,~1,~2,~3,~\ldots)},


    lying between \scriptstyle{n} and \scriptstyle{4n}, where \scriptstyle{\kappa=1} if \scriptstyle{n} is of the form \scriptstyle{8k+3}, and \scriptstyle{\kappa>1} if \scriptstyle{n} is of the form \scriptstyle{8k+7}.

    (6Β·5) \scriptstyle{n\equiv 1\pmod{8}}.

    In this case we have to prove that

    (i) if \scriptstyle{n\geq 33}, there is an infinity of integers which cannot be expressed in the form (5Β·2);

    (ii) if \scriptstyle{n} is \scriptstyle{1}, \scriptstyle{9}, \scriptstyle{17}, or \scriptstyle{25}, there is only a finite number of exceptions.

    In order to prove (i) suppose that \scriptstyle{N=7.4^\lambda}. Then obviously \scriptstyle{u} cannot be zero. But if \scriptstyle{u} is not zero \scriptstyle{u^2} is always of the form \scriptstyle{4^\kappa(8\nu+1)}. Hence

    \scriptstyle{N-nu^2=7.4^\lambda-n.4^\kappa(8\nu+1)}.


    Since \scriptstyle{n\geq 33}, \scriptstyle{\lambda} must be greater than or equal to \scriptstyle{\kappa+2}, to ensure that the right-hand side shall not be negative. Hence

    \scriptstyle{N-nu^2=4^\kappa(8k+7)},