Solution
= Solution
The same recurrence and $2^0\mid u_0$, $2^1\mid u_1$ prove the claim: if $2^{n-1}\mid u_{n-1}$ and $2^{n-2}\mid u_{n-2}$, both terms on the right are divisible by $2^n$.
= Solution
The same recurrence and $2^0\mid u_0$, $2^1\mid u_1$ prove the claim: if $2^{n-1}\mid u_{n-1}$ and $2^{n-2}\mid u_{n-2}$, both terms on the right are divisible by $2^n$.