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

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17
in the form ${\displaystyle \scriptstyle {ax^{2}+by^{2}+cz^{2}+du^{2}}}$
1. (6·4) ${\displaystyle \scriptstyle {n\equiv 3{\pmod {4}}}}$.

There is only a finite number of exceptions. Take

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

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

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

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

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

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 ${\displaystyle \scriptstyle {n}}$ and ${\displaystyle \scriptstyle {4n}}$ for which ${\displaystyle \scriptstyle {\lambda =0}}$. For these ${\displaystyle \scriptstyle {u}}$ must be ${\displaystyle \scriptstyle {1}}$, and

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

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

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

The exceptions required are therefore those of the numbers
 ${\displaystyle \scriptstyle {n+4^{\kappa }(8\nu +7)}}$ (6·41)
which lie between ${\displaystyle \scriptstyle {n}}$ and ${\displaystyle \scriptstyle {4n}}$ and are of the form
 ${\displaystyle \scriptstyle {8\mu +7}}$ (6·42).

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

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

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

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

In this case we have to prove that

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

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

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

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

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

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