There has been a discussion of this problem here: http://www.artofproblemsolving.com/Forum/viewtopic.php?f=39&t=286916

my solution: Because $P(x)\mid Q(x)$ there is a polynomial $R(x)$ such that $Q(x)=P(x)R(x)$ note that because the coefficients of $P(x),Q(x)$ are $1$ or $2002$ we have $Q(x)=P(x)R(x)\longrightarrow Q(x)\equiv P(x)R(x)\pmod{3}\longrightarrow \frac{x^{q+1}-1}{x-1}\equiv \frac{x^{p+1}}{x-1}R(x)\longrightarrow x^{p+1}-1\mid x^{q+1}-1\longrightarrow p+1\mid q+1$ DONE

Essentially the same as that posted above. (An alternate finish by using some theory of algebraic closures) Write the relation $Q=P\cdot R$ in $F_3[X]$. Clearly, we have $X^{p+1}-1 \mid X^{q+1}-1$ since $2002 \equiv 1 \bmod 3$ in $F_3[X]$. Consider the algebraic closure of $F_3$ and call it $\bar{F_3}$. Let us consider the irreducible factor $\phi_{p+1}(X)$ (cyclotomic polynomial of order $p+1$) of $X^{p+1}-1$. Let $\gamma$ be a root of $\phi_{p+1}$ in $\bar{F_3}$. Plugging in $\gamma$ gives that $X^{p+1}-1$ vanishes and $\gamma^{q+1}=1$. However, being a root of the cyclotomic polynomial, which is the minimal polynomial of $\gamma$, we see that it has order $p+1$ and so, $p+1 \mid q+1$. The result follows.

anantmudgal09 wrote: Essentially the same as that posted above. (An alternate finish by using some theory of algebraic closures) Write the relation $Q=P\cdot R$ in $F_3[X]$. Clearly, we have $X^{p+1}-1 \mid X^{q+1}-1$ since $2002 \equiv 1 \bmod 3$ in $F_3[X]$. Consider the algebraic closure of $F_3$ and call it $\bar{F_3}$. Let us consider the irreducible factor $\phi_{p+1}(X)$ (cyclotomic polynomial of order $p+1$) of $X^{p+1}-1$. Let $\gamma$ be a root of $\phi_{p+1}$ in $\bar{F_3}$. Plugging in $\gamma$ gives that $X^{p+1}-1$ vanishes and $\gamma^{q+1}=1$. However, being a root of the cyclotomic polynomial, which is the minimal polynomial of $\gamma$, we see that it has order $p+1$ and so, $p+1 \mid q+1$. The result follows. Any one have solution for this,i wonder thevabove solution is not true:why $x^{p+1}-1|x^{q+1}-1$

andria wrote: $\frac{x^{q+1}-1}{x-1}\equiv \frac{x^{p+1}-1}{x-1}R(x)\longrightarrow x^{p+1}-1\mid x^{q+1}-1$ Can you explain this part $?$

Please help

We will work here on $\mathbb F_{23}$ becuase sadly $69$ is not a prime :c. So on $\mathbb F_{23}$ we have that $P(x)=\frac{x^{p+1}-1}{x-1}$ and $Q(x)=\frac{x^{q+1}-1}{x-1}$ and since we said $P(x) \mid Q(x)$ we have that $(x-1)P(x) \mid (x-1)Q(x)$ and now transforming this on $\mathbb F_{23}$ we have that $x^{p+1}-1 \mid x^{q+1}-1$ meaning that $p+1 \mid q+1$ thus we are done

Taking modun 2001 and do the same as here