# Page:On the expression of a number in the form ππ₯Β²+ππ¦Β²+ππ§Β²+ππ’Β².djvu/7

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

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)β$\scriptstyle{n\equiv 1\pmod{8}}$.

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

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