Solve the equation:
\(8f(m + 1) - f(m) = 2\)
where \(f(m) = m(m + 1)\) when \(m\) is even, and \(f(m) = m + 3\) when \(m\) is odd.
Since \(m\) is even, we use: \(f(m) = m(m + 1)\)
and \(m + 1\) is odd, so: \(f(m + 1) = (m + 1) + 3 = m + 4\)
Substitute into the original equation:
Set equal to 2:
Solve the quadratic:
\[(m - 10)(m + 3) = 0 \Rightarrow m = 10 \text{ or } m = -3\]Only the positive integer solution is: \(\boxed{m = 10}\)
Then \(f(m) = m + 3\) and \(f(m + 1) = (m + 1)(m + 2)\) since \(m + 1\) is even.
Substituting into the equation:
\[8f(m + 1) - f(m) = 8(m + 1)(m + 2) - (m + 3)\]
Set equal to 2:
Set equal to 2:
\[8m^2 + 23m + 13 = 2 \Rightarrow 8m^2 + 23m + 11 = 0\]The discriminant is:
\[D = 23^2 - 4 \cdot 8 \cdot 11 = 529 - 352 = 177\]Since the discriminant is not a perfect square, there is no integer solution.
The only valid solution is when \(m = 10\).
Final Answer: \(\boxed{m = 10}\)
For any natural number $k$, let $a_k = 3^k$. The smallest natural number $m$ for which \[ (a_1)^1 \times (a_2)^2 \times \dots \times (a_{20})^{20} \;<\; a_{21} \times a_{22} \times \dots \times a_{20+m} \] is: