Maybe there's another proof but I would post mine first. $$f(x)^4+2f(x)+2=(x^4+2x^2+2)g(x)+3h(x)$$ Term $h(x)$ simply means that we would like to consider coefficient modulo 3. So the problem is equivalent to "Find $f(x), g(x) \in \mathbb{Z}_3 [x]$ satisfying $f(x)^4+2f(x)+2=(x^4+2x^2+2)g(x)$". Again, the problem is equivalent to "Find $f(x) \in \mathbb{Z}_3[x]$ such that $f(x)^4+2f(x)+2$ is divisible by $x^4+2x^2+2$. (Note that when considering divisibility the coefficient ring is $\mathbb{Z}_3$ not $\mathbb{Z}$ or $\mathbb{Q}$.) Take field extension $\mathbb{F}$ of $\mathbb{Z}_3$ that splits $x^4+2x^2+2$. Since $x^4+2x^2+2$ is irreducible in $\mathbb{Z}_3[x]$, this simply means that we add $\omega$ satisfying $\omega^4 +2 \omega ^2 + 2=0$. $\mathbb{F}$ is the set of $a\omega^3 +b \omega^2 + c\omega +d$ with $a,b,c,d \in \mathbb{Z}_3$. The multiplication can be also defined by using $\omega^4 = - 2 \omega^2 -2 = \omega^2 +1$. $f(x)^4+2f(x)+2$ is divisible by $x^4+2x^2+2$ w.r.t. $\mathbb{Z}_3$ if and only if $f(\omega)^4 + 2f(\omega) + 2 = 0$ in $\mathbb{F}$. When $f(\omega)=a\omega^3 +b\omega^2 +c\omega +d$, $f(x)-(ax^3+bx^2+cx+d)$ is divisible by $x-\omega$. Since its coefficients are in $\mathbb{Z}_3$, $f(x)-(ax^3+bx^2+cx+d)$ is divisible by $x^4+2x^2+2$. Therefore, $f(x)=(ax^3+bx^2+cx+d)+(x^4+2x^2+2)q(x)$ for some $q(x) \in \mathbb{Z}_3[x]$. Since we only want to find polynomial with minimal degree, $q(x)=0$ for our case. In conclusion, the problem is equivalent to find $y=a\omega^3 +b\omega^2 +c\omega +d$ with $y^4+2y+2=0$ with $\omega^4=\omega^2+1$. Well.. I don't know there is some fancy way to find this.. I just plugged that in divide cases and found four roots with $$(a,b,c,d)=(1,1,1,1), (1,2,2,2), (2,1,2,1), (2,2,1,2)$$so therefore the answer is $f(x)=x^3+x^2+x+1, x^3+2x^2+2x+2, 2x^3+x^2+2x+1, 2x^3+2x^2+x+2$. This is not so horrible.. by following two features 1. When calculating $(a\omega^3 +b\omega^2 +c\omega +d)^4$, you only need to consider terms with coefficient in forms of $t^4$ or $t^3 s$. Otherwise, the coefficients are divisible by 3, so just 0 in $\mathbb{F}$. 2. $\omega^8 = (\omega^4)^2 = (\omega^2 +1)^2 = \omega^4 + 2\omega^2 + 1 = 2$. Still I don't this proof is appropriate for MO since it requires student to be used to field theories in abstract algebra. If someone find better way (maybe there is clues to find roots without harsh calculation), please post your solution

bumjoooon wrote: Still I don't this proof is appropriate for MO since it requires students to be used to field theories in abstract algebra. If someone finds a better way (maybe there are clues to find roots without harsh calculation), please post the solution In fact, one could avoid the field extension part by considering the remainder $f(x) \pmod {x^4+2x^2+2}$ in $\mathbb{Z}[x]$, which exists since $x^4+2x^2+2$ is monic. I cannot, however, think of a way to avoid the massive casework though

lminsl wrote: bumjoooon wrote: Still I don't this proof is appropriate for MO since it requires students to be used to field theories in abstract algebra. If someone finds a better way (maybe there are clues to find roots without harsh calculation), please post the solution In fact, one could avoid the field extension part by considering the remainder $f(x) \pmod {x^4+2x^2+2}$ in $\mathbb{Z}[x]$, which exists since $x^4+2x^2+2$ is monic. However, I cannot think a way to avoid the massive casework though Yes, that would be better. Since I first reduce to the $\mathbb{Z}_3[x]$, I thought it would be also annoying to say something about well-definedness of remainder in $\mathbb{Z}_3[x]$. But if we first consider remainder of $f(x)$ in $\mathbb{Z}[x]$, it must be better to write. Good