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

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(6Β·4) n\equiv 3\pmod{4}.

There is only a finite number of exceptions. Take

N=4^\lambda(8\mu+7).
If \lambda\geq 1, take u=1. Then

N-nu^2\equiv 1\ \mathit{or}\ 5\pmod{8}.
If \lambda=0, take u=2. Then

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 n and 4n for which \lambda=0. For these u must be 1, and

N-nu^2\equiv 0\pmod{4}.
But the numbers which are multiples of 4 and which cannot be expressed in the form x^2+y^2+z^2 are the numbers

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

n+4^\kappa(8\nu+7) (6Β·41)
which lie between n and 4n and are of the form

8\mu+7 (6Β·42).


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

n+4^\kappa(8\nu+7),\quad(\nu=0,1,2,3,\ldots),
lying between n and 4n, where \kappa=1 if n is of the form 8k+3, and \kappa>1 if n is of the form 8k+7.

(6Β·5) n\equiv 1\pmod{8}.

In this case we have to prove that

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

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

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

N-nu^2=7.4^\lambda-n.4^\kappa(8\nu+1).
Since n\geq 33, \lambda must be greater than or equal to \kappa+2, to ensure that the right-hand side shall not be negative. Hence

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