Problem 2529

a. We show that the pair $(1, -51^{2})$ is $51$-good but not very good. Let $P(x) = x^{3} - 51^{2} x$. Since $P(51) = P(0)$, the pair $(1, -51^{2})$ is not $n$-good for any positive integer that does not divide $51$. Therefore, $(1, -51^{2})$ is not very good.

On the other hand, if $P(m) \equiv P(k) \pmod{51}$, then $m^{3} \equiv k^{3} \pmod{51}$. By Fermat's theorem, from this we obtain
$$m \equiv m^{3} \equiv k^{3} \equiv k \quad (\bmod 3) \quad \text{and} \quad m \equiv m^{33} \equiv k^{33} \equiv k \quad (\bmod 17).$$
Hence we have $m \equiv k \pmod{51}$. Therefore $(1, -51^{2})$ is $51$-good.

b. We will show that if a pair $(a, b)$ is $2010$-good then $(a, b)$ is $67^{i}$-good for all positive integer $i$.

Claim 1. If $(a, b)$ is $2010$-good then $(a, b)$ is $67$-good.

Proof. Assume that $P(m) \equiv P(k) \pmod{67}$. Since $67$ and $30$ are coprime, there exist integers $m'$ and $k'$ such that $k' \equiv k \pmod{67}$, $k' \equiv 0 \pmod{30}$, and $m' \equiv m \pmod{67}$, $m' \equiv 0 \pmod{30}$. Then we have $P(m') \equiv P(0) \equiv P(k') \pmod{30}$ and $P(m') \equiv P(m) \equiv P(k) \equiv P(k') \pmod{67}$, hence $P(m') \equiv P(k') \pmod{2010}$. This implies $m' \equiv k' \pmod{2010}$ as $(a, b)$ is $2010$-good. It follows that $m \equiv m' \equiv k' \equiv k \pmod{67}$. Therefore, $(a, b)$ is $67$-good.

Claim 2. If $(a, b)$ is $67$-good then $67 \mid a$.

Proof. Suppose that $67 \nmid a$. Consider the sets $\{a t^{2} \pmod{67}: 0 \leq t \leq 33\}$ and $\{-3 a s^{2} - b \pmod{67}: 0 \leq s \leq 33\}$. Since $a \not\equiv 0 \pmod{67}$, each of these sets has $34$ elements. Hence they have at least one element in common. If $a t^{2} \equiv -3 a s^{2} - b \pmod{67}$ then for $m = t \pm s$, $k = \mp 2s$ we have
$$\begin{aligned} P(m) - P(k) = a(m^{3} - k^{3}) + b(m - k) & = (m - k)\left(a(m^{2} + m k + k^{2}) + b\right) \\ & = (t \pm 3s)\left(a t^{2} + 3 a s^{2} + b\right) \equiv 0 \quad (\bmod 67) \end{aligned}$$
Since $(a, b)$ is $67$-good, we must have $m \equiv k \pmod{67}$ in both cases, that is, $t \equiv 3s \pmod{67}$ and $t \equiv -3s \pmod{67}$. This means $t \equiv s \equiv 0 \pmod{67}$ and $b \equiv -3 a s^{2} - a t^{2} \equiv 0 \pmod{67}$. But then $67 \mid P(7) - P(2) = 67 \cdot 5 a + 5 b$ and $67 \nmid 7 - 2$, contradicting that $(a, b)$ is $67$-good.

Claim 3. If $(a, b)$ is $2010$-good then $(a, b)$ is $67^{i}$-good for all $i \geq 1$.

Proof. By Claim 2, we have $67 \mid a$. If $67 \mid b$, then $P(x) \equiv P(0) \pmod{67}$ for all $x$, contradicting that $(a, b)$ is $67$-good. Hence, $67 \nmid b$.

Suppose that $67^{i} \mid P(m) - P(k) = (m - k)\left(a(m^{2} + m k + k^{2}) + b\right)$. Since $67 \mid a$ and $67 \nmid b$, the second factor $a(m^{2} + m k + k^{2}) + b$ is coprime to $67$ and hence $67^{i} \mid m - k$. Therefore, $(a, b)$ is $67^{i}$-good.