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