18.9. (Yugoslavia, 74). For each number $n \in \mathbf{N}$, find the greatest value that the product of natural numbers with a fixed sum $n$ can take.

18.10\%. (England, 81). Find the smallest value of the quantity $\left|12^{m}-5^{n}\right|$ for $m, n \in \mathbf{N}$.

## § 19. Various Properties of Functions

## (see Appendix G: definitions 2, 3, 6;

 theorems $27,28,30,33,34$ )19.1. (GDR, 83). Given two positive numbers $x_{1}$, $x_{2}$, and the function $f: \mathbf{R} \rightarrow \mathbf{R}$ defined by the formula

$$
f(x)=a x^{4}+b x^{2}+c \text{ (where } a, b, c \in \mathbf{R}, a \neq 0)
$$

it is known that $f(0)=f\left(x_{1}\right)=1$ and $f^{\prime}\left(x_{2}\right)=0$. Find the coefficients $a, b, c$.
18.9. Note that any fixed number $n \in N$ can be decomposed into a sum of natural numbers in only a finite number of ways. 298

Therefore, among these representations, there will be one (perhaps not unique) decomposition $n=m_{1}+m_{2}+\ldots+m_{k}$, where $m_{1} \leqslant m_{2} \leqslant \ldots \leqslant m_{k}$, for which the product $m_{1} m_{2} \ldots m_{k}$ takes the maximum value $f(n)$, and no number $m_{i}(i=1,2, \ldots, k)$ equals 4 (any four can be replaced by two twos from the beginning, without changing either the sum or the product of the numbers). Then each number $m_{i}$ does not exceed 4, because if $m_{k}>4$, then the product of the numbers $m_{1}, m_{2}, \ldots, m_{k}-2,2$ will be greater than before (since $\left(m_{k}-2\right) 2 > m_{k}$), while their sum remains $n$. For $n=1$, we have the only decomposition, so $f(1)=1$. Further, for $n>1$, no number $m_{i}$ equals 1, because if $m_{i}=1$, then the product of the numbers $m_{1}+m_{2}, m_{3}, \ldots, m_{k}$ will be greater than before (since $m_{1}+m_{2} > m_{1} m_{2}$), while the sum equals $n$. Finally, among the numbers $m_{i}$, there cannot be three (or more) twos, because if $m_{1}=m_{2}=m_{3}=2$, then the product of the numbers $3,3, m_{4}, \ldots, m_{k}$ will be greater than before (since $3^{2} > 2^{3}$) with the same sum. Thus, if $n>1$, then each number $m_{i}$ equals 3, except, perhaps, one or two numbers equal to 2. These conditions uniquely determine the set of numbers $m_{i}$ for each value of $n>1$. Therefore, for the desired value $f(n)$, we have

$$
f(1)=1, f(3 l)=3^{l}, f(3 l-1)=2 \cdot 3^{l-1}
$$

and

$$
f(3 l+1)=4 \cdot 3^{l-1} \quad \text { for } \quad l \in N .
$$