Let $n \geqslant 2$, determine the remainder of the Euclidean division of $X^{n}$ by $X^{2}-4 X+3$.

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Note that $X^{2}-4 X+3=(X-1)(X-3)$. Let $Q_{n}$ and $R_{n}$ be the quotient and the remainder of $X^{n}$ when divided by $(X-1)(X-3)$, so we have $X^{n}=Q_{n}(X-1)(X-3)+R_{n}$. By evaluating at 1 and 3, we get $R_{n}(1)=1^{n}=1$ and $R_{n}(3)=3^{n}$.

Furthermore, the degree of $R_{n}$ is strictly less than the degree of $(X-3)(X-1)$, which is 2. Therefore, we can write $R_{n}$ in the form $a_{n} X+b_{n}$. Thus, we have $3 a_{n}+b_{n}=3^{n}$ and $a_{n}+b_{n}=1$. By subtracting these equations, we obtain $a_{n}=\frac{3^{n}-1}{2}$, and then $b_{n}=1-a_{n}=\frac{3-3^{n}}{2}$, so $R_{n}=\frac{3^{n}-1}{2} X+\frac{3-3^{n}}{2}$.