Note that we're actually being asked to prove that if there exists an integer polynomial $f$ such that $$p(x)f(x)-x^6-x^4-x^2-1$$has all coefficients divisible by $3$. Let $p(x)=ax^2+bx+c$, the condition that $p(x)$ is nonzero $\mod3$ for all $x$ is equivalent to \begin{align*} & c\not\equiv 0 \pmod 3 \\ & a+b+c \not\equiv 0 \pmod 3 \\ & a+2b+c \not\equiv 0 \pmod 3 \end{align*}for wich the only solutions modulo $3$ are $ (a,b,c)=\pm \{(2,2,1),(2,2,1),(1,0,1),(0,0,1)\}$ For now assume $a=0$ and we can cosider only the plus case of the given solutions above because we can take $f \to -f$ and transform it to the plus sign if we have the negative ones. Note that $\deg f\geq 4$ and for $\deg f >4$ all coefficients near some $x^k,k>4$ should be all $0 \pmod 3$, thus we can consider $deg f=4$. Let $f(x)=r_4x^4+r_3x^3+r_2x^2+r_1x+r_0$ we have that \begin{align*} p(x)q(x)-x^6-x^4-x^2-1=& x^6(ar_4-1)+x^5(ar_3+br_4)+x^4(ar_2+br_3+cr_4-1) \\ &+x^3(ar_1+br_2+cr_3)+x^2(ar_0+br_1+cr_2-1)+x(br_0+cr_1)+cr_0-1 \end{align*}Now, we have the cases: if $(a,b,c)=(1,0,1)$ then it is easy to see that $r_4=r_0=1, r_1=r_2=r_3=0$ works. if $(a,b,c)=(2,1,1)$ we get that $r_4=r_3=r_1=2, r_0=1, r_2=0$ works. if $(a,b,c)=(2,2,1)$ we get that $r_4=2, r_3=r_1=r_0=1, r_2=0$ works. Note that the above solutions are all taken $\pmod 3$. Now if $a=0$ this would imply $b=0,c=1$ then take $f(x)=x^6+x^4+x^2+1$.

XbenX wrote: Note that we're actually being asked to prove that if there exists an integer polynomial $f$ such that $$p(x)f(x)-x^6-x^4-x^2-1$$has all coefficients divisible by $3$. Let $p(x)=ax^2+bx+c$, the condition that $p(x)$ is nonzero $\mod3$ for all $x$ is equivalent to \begin{align*} & c\not\equiv 0 \pmod 3 \\ & a+b+c \not\equiv 0 \pmod 3 \\ & a+2b+c \not\equiv 0 \pmod 3 \end{align*}for wich the only solutions modulo $3$ are $ (a,b,c)=\pm \{(2,2,1),(2,2,1),(1,0,1),(0,0,1)\}$ For now assume $a=0$ and we can cosider only the plus case of the given solutions above because we can take $f \to -f$ and transform it to the plus sign if we have the negative ones. Note that $\deg f\geq 4$ and for $\deg f >4$ all coefficients near some $x^k,k>4$ should be all $0 \pmod 3$, thus we can consider $deg f=4$. Let $f(x)=r_4x^4+r_3x^3+r_2x^2+r_1x+r_0$ we have that \begin{align*} p(x)q(x)-x^6-x^4-x^2-1=& x^6(ar_4-1)+x^5(ar_3+br_4)+x^4(ar_2+br_3+cr_4-1) \\ &+x^3(ar_1+br_2+cr_3)+x^2(ar_0+br_1+cr_2-1)+x(br_0+cr_1)+cr_0-1 \end{align*}Now, we have the cases: if $(a,b,c)=(1,0,1)$ then it is easy to see that $r_4=r_0=1, r_1=r_2=r_3=0$ works. if $(a,b,c)=(2,1,1)$ we get that $r_4=r_3=r_1=2, r_0=1, r_2=0$ works. if $(a,b,c)=(2,2,1)$ we get that $r_4=2, r_3=r_1=r_0=1, r_2=0$ works. Note that the above solutions are all taken $\pmod 3$. Now if $a=0$ this would imply $b=0,c=1$ then take $f(x)=x^6+x^4+x^2+1$. $a+2b+c \not\equiv 0 \pmod 3 $ How did you get this?

consider $$P(2)$$& according to that $$4\equiv 1 \pmod 3 \\$$