<Proof>
\(\cdots\) (omitted) \(\cdots\)
Since \(m\) and \(n\) are integers and \(3m^2=n^2+1\),
\(n^2+1\) has to be a multiple of \(3\).
One of the following holds for some integer \(k\).
If \(n=3k\), then
\(n^2=(3k)^2=9k^2=3(3k^2).\)
If \(n=3k+1\), then
\(\begin{align} n^2 &=(3k+1)^2=9k^2+6k+1\\ &=3(3k^2+2k)+1.
\end{align}\)
If \(n=3k+2\), then
\(\begin{align} n^2 &=(3k+2)^2=9k^2+12k+4\\ &=3(3k^2+4k+1)+1.
\end{align}\)
Therefore, in the division of \(n^2\) by \(3\), the remainder is
either \(0\) or \(1\).
It follows that in the division of \(n^2+1\) by \(3\), the remainder
is either \(1\) or \(2\).
\(\cdots\) (omitted) \(\cdots\)
What is appropriate for \(\fbox{\(\;(\alpha)\;\)}\) above?