SAUDI ARABIAN IMO Booklet 2023

We have the general formula of the given sequence as $$u_n=\frac{(1+\sqrt{3}{)}^n-(1-\sqrt{3}{)}^n}{2\sqrt{3}}.$$ Considering the periodicity of the remainder when divided by $3$ of the given sequence, we have $0,1,2,0,1,2,\dots$ So obviously $u_{3k+1}$ and $u_{3k+2}$ are not divisible by $3$, and then $v_3(u_{3k+1})=v_3(u_{3k+2})=0$.

Next, consider $n=3k$ with $k\in {\mathbb{Z}}^{+}$. We have $$\begin{array}{rl}{u}_{3k}& =\frac{(1+\sqrt{3}{)}^{3k}-(1-\sqrt{3}{)}^{3k}}{2\sqrt{3}}\\ & =\frac{[(1+\sqrt{3}{)}^k-(1-\sqrt{3}{)}^k][(4+2\sqrt{3}{)}^k+(4-2\sqrt{3}{)}^k+(-2{)}^k]}{2\sqrt{3}}\\ & =u_k[(4+2\sqrt{3}{)}^k+(4-2\sqrt{3}{)}^k+(-2{)}^k].\end{array}$$

Put $a_k=(4+2\sqrt{3}{)}^k+(4-2\sqrt{3}{)}^k+(-2{)}^k$ then it is easy to check that $a_0=3$, $a_1=6$, $a_2=60$ and $$a_{n+3}=6a_{n+2}+12a_{n+1}-8a_n,\forall n\ge 0.$$ Notice that $a_0,a_1,a_2$ are all divisible by $3$ so $3|a_n$ for all $n$. Thus $9|6a_{n+2}+12a_{n+1}$ and followed by $a_{n+3}\equiv -8a_n\equiv a_n(\mathrm{mod}9)$. On the other hand, the first three terms of the sequence are not divisible by $9$, so the same applies to all terms of the sequence. From that we have $v_3(a_n)=1$ for all $n$. So $$v_3(u_{3k})=v_3(u_k)+v_3(a_k)=1+v_3(u_k).$$ From here it is easy to see that if we put $n=3^{t}m$ with $\gcd (m,3)=1$ and $t\in {\mathbb{Z}}^{+}$ then $$v_3(u_n)=v_3(u_{3^{t-1}m})+1=\cdots =v_3(u_m)+t=t.$$