Recurring sequence and congruences

Question

Let's define a sequence $(u_n)$ by the formula $u_0 = 3$ and $u_{n+1} = u_n^2 -u_n$.

Is it true (and if so how do you prove it) that $u_n$ is never $0$ and has no square factors (i.e. $u_n \neq 0$ in $\mathbf{Z}/p^2\mathbf{Z}$ for all primes $p$) ?

Answer 1

Incomplete solution:

Modulo $4$ the sequence is $3,2,2,2,\ldots$. Modulo $9$ it is $3,6,3,6,\ldots$. Modulo $25$ it is $3,6,-5,5,-5,5,\ldots$. Now fix a prime $p\ge 7$. By definition $u_{n+1}=u_n(u_n-1)$. Hence $p\mid u_{n+1}$ if and only if $p\mid u_n$ or $p\mid u_n-1$ (note that they are coprime). In particular, let $k\ge 2$ be the smallest integer such that $p\mid u_k$ (if it exists). Since it is the smallest one, then $p \nmid u_{k-1}$. This means that $ u_{k-2}(u_{k-2}-1)=u_{k-1}\equiv 1\bmod{p}, $ which is equivalent to $$ (2u_{k-2}-1)^2 \equiv 5 \bmod{p}. $$ (This has a solution if and only if $p\equiv \pm 1\bmod{5}$.)

Unproved claim: $p^2 \nmid u_k$.

Assuming the above claim is correct, then $u_k$, being divisible by $p$, has to be equal to $mp$, for some integer $m$ not divisible by $p$. Then, it is easy to check that $u_{k+n}=(-1)^n u_k \bmod{p^2}$, for all $n\ge 0$, hence $0$ will never appear in the sequence.