Page:On the expression of a number in the form 𝑎𝑥²+𝑏𝑦²+𝑐𝑧²+𝑑𝑢².djvu/7

6. (6·4) $\scriptstyle {n\equiv 3{\pmod {4}}}$.

   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

   $\scriptstyle {n+4^{\kappa }(8\nu +7)}$

   (6·41)

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

   $\scriptstyle {8\mu +7}$

   (6·42).

   
   

   Now in order that (6·41) may be of the form (6·42), $\scriptstyle {\kappa }$ 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 (5·2), 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 $\scriptstyle {8k+7}$.

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

   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 (5·2);

   (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

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