Since mapping $x\to x^3+1\pmod p$ is injectivr, we get the sequence $a_n\pmod p$ is periodic. This means for some $n$, $a_{n+1}\equiv a_1\equiv 1\pmod p$ $\implies a_n\equiv 0\pmod p$.

MarkBcc168 wrote: Since mapping $x\to x^3+1\pmod p$ is injectivr Why?

Bandera wrote: MarkBcc168 wrote: Since mapping $x\to x^3+1\pmod p$ is injectivr Why? If $x\neq y$ modulo $p$, then let $t=\frac{x}{y}$. $$x^3+1=y^3+1 \Longleftrightarrow (x-y)(x^2+xy+y^2)=0 \Longleftrightarrow t^2+t+1=0 \Longleftrightarrow (2t+1)^2=-3 \pmod p$$But $\left(\frac{-3}{p}\right)=-1$.

rmtf1111 wrote: $\left(\frac{-3}{p}\right)=-1$. Wow! Law of quadratic reciprocity in a school Olympiad

$x^{p-1}=y^{p-1}=1\left(\operatorname{mod}p\right)$ $\left(x^3\right)^{\frac{p-2}{3}}\cdot x=\left(y^3\right)^{\frac{p-2}{3}}\cdot y\left(\operatorname{mod}p\right)$ If $x^3=y^3\left(\operatorname{mod}p\right)$ then we can conclude that $x=y\left(\operatorname{mod}p\right)$

consider the following set. $${a_1,a_2,...,a_p,a_{p+1}}$$exist $m$ and $n$ $$a_m\equiv a_n (modp)$$$$p\equiv 2(mod3)$$$$a_{m-1}\equiv a_{n-1} (modp)$$$$\cdot$$$$\cdot$$$$\cdot$$$$a_{m-n} \equiv a_1 \equiv 1(modp)$$$$(a_{m-n-1})^3+1 \equiv 1(modp)$$$$p|a_{m-n-1}$$

More generally, if $x^k\equiv y^k\pmod p$ where $k$ is coprime with $p-1$, there exists a positive integer $t$ such that $p-1\mid kt-1$. Then, $x\equiv x^{kt}\equiv y^{kt}\equiv y\pmod p$.