Problem 9.3. Find the number of positive integers $a$ less than 2003 , for which there exists a positive integer $n$ such that $3^{2003}$ divides $n^{3}+a$.

Emil Kolev, Nikolai Nikolov
9.3. We shall prove that the desired number have one of the forms $9 k \pm 1$, $3^{3}(9 k \pm 1)$ or $3^{6}(9 k \pm 1)$.

Suppose that 3 does not divide $a$. Since $n^{3} \equiv 0, \pm 1(\bmod 9)$, then $a \equiv \pm 1$ $(\bmod 9)$.

Conversely, let $a \equiv \pm 1(\bmod 9)$. Since 9 divides $1^{3}-1$ and $2^{3}+1$, then there is $n_{0}$ such $n_{0}^{3}+a=3^{s} t$, where $s \geq 2$ and $t$ is not divisible by 3 . We shall prove that if $n_{1}=n_{0}+2.3^{s-1} t$, then $3^{s+1}$ divides $n_{1}^{3}+a$. We have that

$$
\left(n_{0}+2.3^{s-1} t\right)^{3}+a=3^{s} t\left(2 n_{0}^{2}+1\right)+4 n_{0} 3^{2 s-1} t^{2}+8.3^{3 s-3} t^{3}
$$

Since 3 does not divide $n_{0}$, then $2 n_{0}^{2}+1$ is divisible by 3 . Moreover, $2 s-1 \geq s+1$ and $3 s-3 \geq s+1$. Hence $n_{1}^{3}+a$ is divisible by $3^{s+1}$ but 3 does not divide $n_{1}$. Repeating the same argument, we get a positive integer $n_{p}$ such that $3^{2003}$ divides $n_{p}^{3}+a$.

Let now 3 divides $a<2003$. Then $a=3^{s} b$, where $s \leq 6$. Hence $n$ is divisible by 3 , i.e., $n=3^{p} n_{0}$, where $p \geq 1$ and 3 does not divide $n_{0}$. If $p \geq 3$, then $3^{9}$ divides $n^{3}$ and does not divide $a$ which implies that $3^{2003}$ does not divide $n^{3}+a$. Hence $p=1$ or $p=2$ and it is easy to see that $s=3$ or $s=6$, respectively.

In the first case we get that $3^{2000}$ divides $n_{0}^{3}+b$, where 3 does not divide $b$ and $27 b<2003$. It follows as above that $b \equiv \pm 1(\bmod 9)$.

In the second case we get similarly that $3^{1997}$ divides $n_{0}^{3}+b$, where $729 b<$ 2003 and $b \equiv \pm 1(\bmod 9)$.

The number of the positive integers $b \equiv \pm 1(\bmod 9)$ such that $b<2003$, $27 b<2003$ or $729 b<2003$ equals $2.222+1=445,2.8+1=17$ or 1 , respectively. Hence the desired number is equal to $445+17+1=463$.