
(6Β·4)β.
There is only a finite number of exceptions. Take
.
If , take . Then.
If , take . Then.
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 and for which . For these must be , and
.
But the numbers which are multiples of and which cannot be expressed in the form are the numbers.
The exceptions required are therefore those of the numbers(6Β·41)
which lie between and and are of the form(6Β·42).
Now in order that (6Β·41) may be of the form (6Β·42), must be if is of the form and may have any of the values if is of the form . Thus the only numbers which cannot be expressed in the form (5Β·2), in this case, are those of the form less than and those of the form,
lying between and , where if is of the form , and if is of the form .(6Β·5)β.
In this case we have to prove that
(i) if , there is an infinity of integers which cannot be expressed in the form (5Β·2);
(ii) if is , , , or , there is only a finite number of exceptions.
In order to prove (i) suppose that . Then obviously cannot be zero. But if is not zero is always of the form . Hence
.
Since , must be greater than or equal to , to ensure that the righthand side shall not be negative. Hence,
Page:On the expression of a number in the form ππ₯Β²+ππ¦Β²+ππ§Β²+ππ’Β².djvu/7
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