## Task 1 - 191241

Determine all pairs $(f(x) ; g(x))$ of polynomials of degree 3

\[
\begin{aligned}
& f(x)=a_{3} x^{3}+a_{2} x^{2}+a_{1} x+a_{0} \\
& g(x)=b_{3} x^{3}+b_{2} x^{2}+b_{1} x+b_{0}
\end{aligned}
\]

whose coefficients $a_{0}, a_{1}, a_{2}, a_{3}, b_{0}, b_{1}, b_{2}, b_{3}$ are real numbers and for which the following conditions are satisfied:

(1) Each of the values that $f(x)$ and $g(x)$ take for $x=1,2,3$ and 4 is one of the numbers 0 and 1.

(2) If $f(1)=0$ or $f(2)=1$, then $g(3)=0$ and $g(4)=1$.

(3) If $f(1)=1$ or $f(4)=1$, then $g(1)=1$ and $g(3)=1$.

(4) If $f(2)=0$ or $f(4)=0$, then $g(2)=0$ and $g(4)=0$.

(5) If $f(3)=1$ or $f(4)=1$, then $g(1)=0$.
Assuming for two polynomials $f(x), g(x)$ the conditions of the problem statement are satisfied.

1. First, it is shown that the function values $f(k), g(k)$ for $k=1,2,3,4$ are uniquely determined by conditions (1) to (5). If $f(4)=1$, then (3) and (5) would lead to the contradiction $g(1)=1$, $g(1)=0$. Therefore, according to (1), $f(4)=0$. From (4), it follows that $g(2)=0, g(4)=0$.

From (2), it follows that neither $f(1)=0$ nor $f(2)=1$ can be true, so according to (1), $f(1)=1, f(2)=0$. From this, (3) yields $g(1)=1, g(3)=1$.

Thus, (5) implies that $f(3)=1$ cannot hold; according to (1), $f(3)=0$. Therefore, we obtain

$$
f(1)=1 ; \quad f(2)=0 ; \quad f(3)=0 ; \quad f(4)=0 ; \quad g(1)=1 ; \quad g(2)=0 ; \quad g(3)=1 ; \quad g(4)=0
$$

2. The equations (6) written out in detail are:

$$
\begin{aligned}
& a_{3}+a_{2}+a_{1}+a_{0}=1 \\
& 8 a_{3}+4 a_{2}+2 a_{1}+a_{0}=0 \\
& 27 a_{3}+9 a_{2}+3 a_{1}+a_{0}=0 \\
& 64 a_{3}+16 a_{2}+4 a_{1}+a_{0}=0 \quad \text { and } \\
& b_{3}+b_{2}+b_{1}+b_{0}=1 \\
& 8 b_{3}+4 b_{2}+2 b_{1}+b_{0}=0 \\
& 27 b_{3}+9 b_{2}+3 b_{1}+b_{0}=1 \\
& 64 b_{3}+16 b_{2}+4 b_{1}+b_{0}=0
\end{aligned}
$$

These are two systems of linear equations for the coefficients, which can be solved, for example, by step-by-step elimination of an unknown. This yields:

$$
\begin{aligned}
& a_{0}=4 ; \quad a_{1}=-\frac{13}{3} ; \quad a_{2}=\frac{3}{2} ; \quad a_{3}=-\frac{1}{6} \quad \text { and } \\
& b_{0}=8 ; \quad b_{1}=-\frac{34}{3} ; \quad b_{2}=5 ; \quad b_{3}=-\frac{2}{3}
\end{aligned}
$$

Therefore, only the polynomials

$$
f(x)=-\frac{1}{6} x^{3}+\frac{3}{2} x^{2}-\frac{13}{3} x+4 \quad \text { and } \quad g(x)=-\frac{2}{3} x^{3}+5 x^{2}-\frac{34}{3} x+8
$$

can satisfy the conditions of the problem statement. That (1) is satisfied can be easily verified by a check. (2) is satisfied, as neither $f(1)=0$ nor $f(2)=1$. (3) is satisfied, as $g(1)=1$ and $g(3)=1$. (4) is satisfied, as $g(2)=0$ and $g(4)=0$, and (5) is satisfied, as neither $f(3)=1$ nor $f(4)=1$.

Therefore, exactly the pair $(f(x), g(x))$ with $f(x), g(x)$ from (7) satisfies all the conditions of the problem statement.

Taken from [3]