Vietnamese Mathematical Olympiad

a) For $a=3,b=-2,c=-1$, we have $u_{n+2}=3u_{n+1}-2u_n-1,\forall n\ge 1$. By induction, one can prove $$u_n=2\alpha -\beta +(\beta -\alpha -1)\cdot 2^{n-1}+n$$ for all $n\ge 1$. Then $u_{2023}=2\alpha -\beta +(\beta -\alpha -1)2^{2022}+2023$. For any $t\in \mathbb{Z}$, choose $\alpha =(2^{2022}-1)t-2021$ and $\beta =t-\alpha -2$. In other words, $2-\beta +\alpha =t$ and $\alpha +2021+(1-2^{2022})t=0$, so we have $$\begin{array}{rl}& 2\alpha -\beta +(\beta -\alpha -1)2^{2022}+2023\\ & =\alpha +(2-\beta +\alpha )+2021+(\beta -\alpha -2)2^{2022}-2^{2022}\\ & =\alpha +t+2021-t\cdot 2^{2022}+2^{2022}\\ & =\alpha +2021+(1-2^{2022})t+2^{2022}=2^{2022}.\end{array}$$ This implies that there exist infinitely many pairs of $(\alpha ,\beta )$ for $u_{2023}=2^{2022}$.

b) Note that $2023=7\times 17^2$. Let $(r_n)$ be the remainder of $(u_n)$ when divided by 2023. Then it is easy to check that $(r_n)$ is the periodic sequence with some period, denote by $T>0$. We consider the following cases:

If there exists $n_0\in {\mathbb{N}}^{\ast }$ such that $7\mid u_{n_0}$ or $17\mid u_{n_0}$. Let consider the first case while the other case does the same. Since $7\mid 2023$, $$\prod _{i=0}^m{u}_{n_0+i}\ne 1(\mathrm{mod}2023),\forall m\ge 1.$$ Hence proposition ii) is not satisfied. For all $l\in {\mathbb{N}}^{\ast }$, choose $m=(2023l-1)T$. Because $u_{n_0+nT}\equiv u_{n_0}(\mathrm{mod}2023)$, and $7\mid 2023$ so $u_{n_0+nT}\equiv u_{n_0}\equiv 0(\mathrm{mod}7)$ for all $n\in {\mathbb{N}}^{\ast }$. Thus the sequence $u_{n_0},u_{n_0+1},\dots ,u_{n_0+(2023l-1)T}$ contains at least 2023 terms which are divisible by 7. Hence, $$\prod _{i=0}^m{u}_{n_0+i}\text{ is divisible by }{7}^{2023}.$$ Therefore, proposition i) is satisfied.

If $7\nmid u_n,17\nmid u_n$ for all $n\in {\mathbb{N}}^{\ast }$, choose $n_0=1$, obviously proposition i) is not satisfied. Otherwise, $\gcd (u_n,2023)=1$, so by Euler's theorem, $$u_n^{\phi (2023)}\equiv 1(\mathrm{mod}2023),\forall n\in {\mathbb{N}}^{\ast }.$$ Set $a=\phi (2023)$, for all $l\in {\mathbb{N}}^{\ast }$, then choose $k=laT$. We will prove that $2023\mid {\prod }_{i=0}^k{u}_{1+i}-1$. Indeed, we have $$u_1\equiv u_{1+T}\equiv \cdots \equiv u_{1+(la-1)T}(\mathrm{mod}2023)$$ $$u_2\equiv u_{2+T}\equiv \cdots \equiv u_{2+(la-1)T}(\mathrm{mod}2023)$$ ...... $$u_T\equiv u_{2T}\equiv \cdots \equiv u_{laT}(\mathrm{mod}2023).$$ Hence $u_i{u}_{i+T}\cdots u_{i+(la-1)T}\equiv u_i^a\equiv 1(\mathrm{mod}2023)$ for all $i=\overline{1,T}$. From this we conclude that ${\prod }_{i=0}^k{u}_{1+i}\equiv 1(\mathrm{mod}2023)$ or $2023\mid {\prod }_{i=0}^k{u}_{1+i}-1$. So proposition ii) is true. $\square$