Solution: we must have $2^n=3k+2$, because $x^3=1$. So ${{2}^{n}}=3k+2\Rightarrow {{(3-1)}^{n}}=3k+2\Leftrightarrow M3+{{(-1)}^{n}}=3k+2\Rightarrow n=2m+1\Rightarrow m\in \left\{ 3,8,...,98 \right\}$ For example if $n=3\Rightarrow {{2}^{3}}=8\Rightarrow {{x}^{8}}+x+1={{\left( {{x}^{3}} \right)}^{2}}{{x}^{2}}+x+1={{x}^{2}}+x+1\vdots {{x}^{2}}+x+1$

Let $\omega$ be a $3$rd root of unity that is not $1,$ we must have $$\omega^{2^{n}}+\omega+1=0 \implies x^2=x^{2^{n}} \implies 2 \equiv 2n \pmod{3} \implies n \equiv 1 \pmod{3}.$$So there are $\boxed{32}$ $n$s that satisfy this, and namely they are $4, 7, 10, ..., 97.$ $\blacksquare$

math31415926535 wrote: Let $\omega$ be a $3$rd root of unity that is not $1,$ we must have $$\omega^{2^{n}}+\omega+1=0 \implies x^2=x^{2^{n}} \implies 2 \equiv 2n \pmod{3} \implies n \equiv 1 \pmod{3}.$$So there are $\boxed{32}$ $n$s that satisfy this, and namely they are $4, 7, 10, ..., 97.$ $\blacksquare$ If $n=4\Rightarrow {{2}^{4}}=16\Rightarrow {{x}^{16}}+x+1={{\left( {{x}^{3}} \right)}^{5}}x+x+1=2x+1$ it is not divisible by ${{x}^{2}}+x+1$ , and the others too!

The.survivor wrote: math31415926535 wrote: Let $\omega$ be a $3$rd root of unity that is not $1,$ we must have $$\omega^{2^{n}}+\omega+1=0 \implies x^2=x^{2^{n}} \implies 2 \equiv 2n \pmod{3} \implies n \equiv 1 \pmod{3}.$$So there are $\boxed{32}$ $n$s that satisfy this, and namely they are $4, 7, 10, ..., 97.$ $\blacksquare$ If $n=4\Rightarrow {{2}^{4}}=16\Rightarrow {{x}^{16}}+x+1={{\left( {{x}^{3}} \right)}^{5}}x+x+1=2x+1$ it is not divisible by ${{x}^{2}}+x+1$ , and the others too! I don't really get what you mean by that

The $4, 7, 10, ..., 97.$ there are not solutions the solutions there are $3,8,...,98$ and proved that your "solution" $n=4$ do not work.

The.survivor wrote: The $4, 7, 10, ..., 97.$ there are not solutions the solutions there are $3,8,...,98$ and proved that your "solution" $n=4$ do not work. then can you tell me where in my sol is flawed?

If $n=4\Rightarrow {{2}^{4}}=16\Rightarrow {{x}^{16}}+x+1={{\left( {{x}^{3}} \right)}^{5}}x+x+1=2x+1$ it is not divisible by ${{x}^{2}}+x+1$ , and the others too!

I know but can you say why my solution is incorrect? Like which step.

$\omega^{2^{n}}+\omega+1=0 \implies x^2=x^{2^{n}} \implies 2 \equiv 2n \pmod{3} \implies n \equiv 1 \pmod{3}.$ Correct: ${{\omega }^{{{2}^{n}}}}+\omega +1=0\Rightarrow {{x}^{2}}={{x}^{{{2}^{n}}}}\Rightarrow {{2}^{n}}\equiv 2\quad \left( \bmod 3 \right)$

o thanks lol, I am pretty dumb

Similar to OTIS Mock AIME 2024/7. All odd $n$ work. The idea is to "mod out" by the polynomial $x^2 + x + 1$ by using $x^3 \equiv 1 \pmod{x^2 + x + 1}$, which shortcuts the polynomial division algorithm. In particular, if $n$ is odd we have \[2^n \equiv 2 \pmod 3 \implies x^{2^n} + x + 1 \equiv x^2 + x + 1 \equiv 0 \pmod{x^2 + x + 1},\]whereas $n$ even gives \[2^n \equiv 1 \pmod 3 \implies x^{2^n} + x + 1 \equiv 2x + 1 \not\equiv 0 \pmod{x^2 + x + 1}.\]So $n$ works if and only if $n$ is odd, which gives a total of $\fbox{49}$ integers ($3, 5, \dots, 99$).