ثابت است $f$ تابع $ \to f(x) = k \to k + k = k \times k $
$\to 2k = {k^2} \to k = 0\,,\,2$
مولفه اول و دوم زوج مرتبها برابراند:
$2{n^2} - 7n + 1 = - f(m) \to $
$\left\{ \begin{gathered}
\xrightarrow{{k = 0}}2{n^2} - 7n + 1 = 0 \hfill \\
\to n = \frac{{7 \pm \sqrt {41} }}{4} \hfill \\
\xrightarrow{{k = 2}}2{n^2} - 7n + 1 = - 2 \hfill \\
\to 2{n^2} - 7n + 3 = 0 \hfill \\
\to n = \frac{1}{2}\, , n = 3 \hfill \\
\end{gathered} \right.$
${m^2} - 4m + 6 = nf(n)$
$\xrightarrow[{n = 3}]{{k = 2}}{m^2} - 4m + 6 = 6$
${m^2} - 4m = 0$
$ \to m = 0\,,\,4\xrightarrow{{m \in \mathbb{N}}}m = 4$
$\left[ {\frac{{mn}}{5}} \right] = \left[ {\frac{{4 \times 3}}{5}} \right] = 2$