(6Β·4)β
.
There is only a finite number of exceptions. Take
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. |
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If
, take
. Then
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. |
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If
, take
. Then
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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
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. |
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But the numbers which are multiples of
and which cannot be expressed in the form
are the numbers
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. |
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The exceptions required are therefore those of the numbers
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 |
(6Β·41) |
which lie between
and
and are of the form
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 |
(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
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, |
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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
In order to prove (i) suppose that
. Then obviously
cannot be zero. But if
is not zero
is always of the form
. Hence
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. |
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Since
,
must be greater than or equal to
, to ensure that the right-hand side shall not be negative. Hence
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