The following is a part of a proof of the statement ‘If
\(x^2+y^2+z^2=1111\), then \(\fbox{\(\;(\alpha)\;\)}\).’
<Proof>
\(\cdots\) (omitted) \(\cdots\)
For the division of integers \(x, y\) and \(z\) by \(8\),
the remainders are either \(0,1,2,3,4,5,6\) or \(7\).
Then, for the division of \(x^2, y^2\) and \(z^2\) by \(8\),
the remainders are either \(0,1\) or \(4\).
Therefore, for the division of \(x^2+y^2+z^2\) by \(8\), the
remainder is either \(0,1,2,3,4,5\) or \(6\). However, the division
of \(1111\) by \(8\) has a remainder of \(7\).
\(\cdots\) (omitted) \(\cdots\)
What is appropriate for \(\fbox{\(\;(\alpha)\;\)}\) above?
[2 points]
- at least one of \(x, y\) and \(z\) is an integer.
- none of \(x, y\) and \(z\) are integers.
-
there is at least one solution such that \(x, y\) and \(z\) are
all integers.
-
there is only one solution such that \(x, y\) and \(z\) are all
integers.
-
there are no solutions such that \(x, y\) and \(z\) are all
integers.