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

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

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$\scriptstyle{N-nu^2\equiv 1\ \mathit{or}\ 5\pmod{8}}$.|undefinedIf $$\scriptstyle{\lambda=0}$$, take $$\scriptstyle{u=2}$$. Then$\scriptstyle{N-nu^2\equiv 3\pmod{8}}$.|undefinedIn 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

$\scriptstyle{N-nu^2\equiv 0\pmod{4}}$.|undefinedBut 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 numberswhich lie between $$\scriptstyle{n}$$ and $$\scriptstyle{4n}$$ and are of the form

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) $\scriptstyle{n\equiv 1\pmod{8}}$.|undefined

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