Polynomial $f(x)$ ($\deg f = 2$) over the field $\Bbb F_5$ takes the value 1 at the point 1, value 2 at the point 3 and value 3 at the point 4. Find $f(x)$.
As I understand, we have 3 variables a,b,c in $f(x)=ax^2+bx+c$ and 3 equations from the task. Thus I have to solve the system of equations such as $a+b+c = 1 (\text{mod }5)$... - am I right?
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How many polynomials of the degree $2$ are there over the prime field $\mathbb F_5$?. There are $100$ (Why?) from which you have to discard $99$ of them which is implicitly proved because the system $$a\cdot1+b\cdot 1+c=1\\a\cdot4+b\cdot 3+c=2\\a\cdot1+b\cdot 4+c=3$$ has exactly one solution which give you the required values of $a,b,c$.
Yes, you're right. But you can also use Lagrange interpolating polynomials.
Edit: the system is $$\begin{bmatrix} a + b + c \\ 4a+3b+c \\ a+4b+c\end{bmatrix} = \begin{bmatrix} 1&1&1\\ 4&3&1 \\ 1&4&1\end{bmatrix} \begin{bmatrix} a \\ b \\c\end{bmatrix}= \begin{bmatrix} 1\\ 2 \\ 3\end{bmatrix}$$
Apply Gaussian elimination as in $\Bbb R$ but taking into account that $5=0 \;\mod 5$...