(a) Let $a = 1$ and $b = -51^2$. We first show that $(a,b)$ is 51-good. If $3 | P(m) - P(k) = (m-k)(a(x^2 + xy + y^2) + b)$, then $3|m-k$ or $3|a(m^2 + mk + k^2) + b$. If $3 | a(m^2 + mk + k^2) + b$, we have $3 | m^2 + mk + k^2 | m^3 - k^3$, so $m \equiv m^3 \equiv k^3 \equiv k \pmod{3}$, so $3 | m-k$ whenever $3 | P(m) - P(k)$. Likewise, if $17| P(m) - P(k)$, we have $17 | m-k$ or $17 | a(m^2 + mk + k^2) + b$. If $17 | a(m^2 + mk + k^2) + b$, then $17 | m^2 + mk + k^2$. If $k \not \equiv 0 \pmod{17}$, we have $(2\frac{m}{k} + 1)^2 \equiv -3 \pmod{17}$, which is impossible since $\left( \frac{-3}{17} \right) = \left( \frac{17}{3} \right) = -1$. Thus, $17 | k$, so $17 | m$, whence $17 | m-k$. Thus, whenever $17 | P(m) - P(k)$, we also have $17 | m - k$. Consequently, if $51 | P(m) - P(k)$, then $51 | m - k$, so $(a,b)$ is 51-good. To show that $(a,b)$ is not very good, observe that $P(51) - P(0) = 0$, whence for all positive integers $n$, $n | P(51) - P(0)$. Consequently, if $(a,b)$ is $n$-good, we must have $n | 51$, so there are only finitely many $n$ for which $(a,b)$ is $n$-good. (b) We claim that if $(a,b)$ is 2010-good, then $67 | a$ and $67 \not | b$. Since $\left(\frac{-3}{67}\right) = 1$, we can find some $x$ with $2 \leq x \leq 66$ such that $x^2 + x + 1 \equiv 0 \pmod{67}$. If $67 | b$, take $m = 30x$ and $k = 30$. Then $P(m) - P(k) = 30(x-1)(30^2(x^2 + x + 1) + b)$ is divisible by $30 \cdot 67 = 2010$. On the other hand, $2010 \not | m - k$, since $0 < m-k = 30(x-1) < 2010$. Thus, $67 \not | b$. Now suppose that $67 \not | a$. We claim that we can find distinct integers $m,k$ such that $0 \leq m,k \leq 66$ and $m^2 + mk + k^2 \equiv \frac{-b}{30^2 a} \pmod{67}$. There are exactly 34 possible values that $u^2 \pmod{67}$ can assume for $0 \leq v \leq 66$, and $34$ possible values that $\frac{-4b}{30^2 a} - 3v^2 \pmod{67}$ can assume for $0 \leq v \leq 66$. Since $34 + 34 > 67$, we can find some $u$ and $v$ such that $u^2 \equiv \frac{-4b}{30^2 a} - 3v^2 \pmod{67}$, i.e., $\frac{u^2 + 3v^2}{4} \equiv \frac{-4b}{30^2 a} \pmod{67}$. Set $k_1 \equiv v \pmod{67}$ and $m_1 \equiv \frac{u-k_1}{2} \pmod{67}$ with $0 \leq m_1, k_1 \leq 66$ to get $\frac{(2m_1+k_1)^2 + 3k_1^2}{4} \equiv \frac{-b}{30^2 a} \pmod{67}$, i.e., $m_1^2 + m_1k_1 + k_1^2 \equiv \frac{-b}{30^2 a} \pmod{67}$. If $m_1 = k_1$, then by Vieta's formula, we have $m_2^2 + m_2k_1 + k_1^2 \equiv \frac{-b}{30^2 a} \pmod{67}$ as well, where $m_2 \equiv -m_1 - k_1 \pmod{67}$ and $0 \leq m_2 \leq 66$. Note that $m_2 \neq k_1$, or else $0 \equiv m_2 + m_1 + k_1 = 3k \pmod{67}$, whence $m_1 = k_1 = 0$, contradicting $67 \not | b$. Thus, if $m_1 \neq k_1$, we can take $(m,k) = (m_1, k_1)$; otherwise, we can take $(m,k) = (m_2, k_1)$. Suppose without loss of generality that $m>k$. We have $P(30m) - P(30k) = 30(m-k)(30^2 a(m^2 + mk + k^2) + b)$. Since $m^2 + mk + k^2 \equiv \frac{-b}{30^2 a} \pmod{67}$, we have $67 | 30^2 a(m^2 + mk + k^2) + b$, whence $2010 | P(30m) - P(30k)$. On the other hand, $0 < 30(m-k) < 2010$, so $2010 \not | 30(m-k)$, a contradiction. We now claim that $(a,b)$ is $67^j$-good for all positive $k$. Indeed, if $67^j | P(m) - P(k) = (m-k)(a(m^2 + mk + k^2) + b)$ , we must have $67 \not | a(m^2 + mk + k^2) + b$ since $67 | a$ but $67 \not | b$, whence $67^j | m-k$, as desired.

Ya your generalization is true.Actually for any integer of the form $n=cp^j$ with $p \equiv 1\pmod{3}$ i think the problem statement is valid.

I don't think so. For any prime $p,$ $p^3 \equiv (-p)^3 \equiv 0 \pmod{p^2},$ so the lemma fails for $c=1$ and $j=2$ (more generally, any even $j$). The generalization is valid for numbers of the form $p_1 p_2 \cdots p_n,$ where the $p_i$'s are distinct primes each congruent to $2 \pmod{3}.$

(a) Let $a=1,b=-3(17)^2$. Clearly it is $51$ good. It is sufficient to see that for any big prime $q$, $(a,b)$ is not $q$-good and that for any prime $q$ and a big $x$, $(a,b)$ is not $q^x$-good. For the second part, take $m=k=17$ mod $q^{x-1}$ but not mod $q^x$. Then $m-k$ isn't divisible by $q^x$ but $(m-k)(a(m^2+mk+k^2)+b))$ is, since $a(m^2+mk+k^2)+b = 3(17)^2-3(17)^2=0$ mod $q$. For the first part assume there are arbitrarily big $q$ that don't work, i.e. $m^2+mk+k^2 = 3(17)^2$ mod $q$ implies $m=k$. Then, $x^2+x(k)+(k^2-3(17)^2)$ doesn't have a solution for $k \neq \pm 17$. This implies $12(17)^2-3k^2$ is not a quadratic residue for $k \neq \pm 17$, and thus $12(17)^2 = 3k_1^2+k_2^2$ mod $q$ is impossible if $k_1 \neq \pm 17$. First notice if $k_2=k_1=k$ then $12(17)^2=4k^2$ is impossible (even for $k= \pm 17$ if $q$ is big enough) and thus $3$ is not a quadratic residue mod $q$. Now divide by $17$ (for $q>17$) and we get $12=3k_1^2+k_2^2$ is impossible mod $q$ if $k_1^2 \neq 1$. Thus in the equation $12=3a+b$ in $\mathbb{Z}_q$, any $a$ that is a non-1 r.c. is paired up with a non-r.c. $b$, and any $a$ that is not a non-1 r.c. is paired up with an r.c. $b$, since there are $(q+1)/2$ r.c. $b$'s and $(q+1)/2$ numbers that are not non-1 r.c.'s. Notice $3$ is a non-r.c. and thus if $b=3$ then it is paired up with a non-1 r.c. $a$, that is, $12=3a+3$ is true for a non-1 r.c. $a$, so $9=3(a_0)^2$ is true for some $a_0$ and thus $3=a_0^2$ is possible, contradicting the fact that 3 is a non-r.c. Thus done. note: r.c.=quadratic residue (b) Let $p=67$. Notice $(a,b)$ is $p$-good. Now let $\Omega=\{m^2+mk+k^2 : m \neq k \}$ in $\mathbb{Z}_p$. We have $aw+b \neq 0$ for any $w \in \Omega$. Let $(m/k)$ have order $3$ mod $p$ (since $3|66$) and then $p|m^2+mk+k^2$ and $m \neq k$ and so $0 \in \Omega$, thus $b \neq 0$. By $m=-1,k=2$ we get $3 \in \Omega$ but by quadratic reciprocity, $\left( \displaystyle\frac{3}{67} \right) = -1$ and thus if $m=-x,k=2x$ then $k \neq m$ (if $x \neq 0$) and $3x^2 \in \Omega$, and $3x^2$ is not an rc. Thus, any non-rc is in $\Omega$. Also notice that if $m=0,k=x \neq 0$ we get that any rc is in $\Omega$. Thus, everyone is in $\Omega$ and the only way to avoid $p | aw+b$ is by $a = 0$ (notice we had $b\neq 0$). Now it is trivial to see $(a,b)$ is $67^{M_{\nu}}$ good for any $M_{\nu}$. Thus done.

A nice number theory problem indeed, and people say that number theory is dead/it is non-feasible to make a nice NT. Part a. Choose $a=1$ and $b=-51^2$. Now, $f(x) \equiv x^3\bmod 51$ and it suffices to prove that we have $51$ cubic residues modulo $51$,which is an easy case bash. Now, we prove that $(1,-51^2)$ is not very good. Indeed, $f(51)=f(0)$ and so we get that for any number $n >51$,$(1,-51^2)$ is not $n$-good. Part b. We claim that $(a,b)$ is a good pair for every $n=67^k$ for all $k \in \mathbb{N}$ Indeed, there cant exist $x,y$ which are incongruent mod $67$ and $67 \mid a(x^2-xy+y^2)+b$. Therefore, if $67^t \mid P(x)-P(y)$ then if $67^t$ does not divide $a(x^2-xy+y^2)+b$ means that we are done. So, $67 \mid 3x^2a+b$. And so $-3ab$ is a quadratic residue mod $67$ giving that $ab$ is a quadratic residue mod $67$. Now, note that $-7$ is a quadratic residue mod $67$ and let $l^2 \equiv -7 \bmod 67$. Now, choosing $x \equiv \frac{y(l-1)}{2}$ we get that $67 \mid a(a(x^2-xy+y^2)+b)$ for a suitable choice of $y$ as $ab$ is a quadratic residue modulo $67$. But, then we have found $x,y$ which violate the hypothesis proving our desired claim.

A convoluted alternative, for part $b)$. $b)$ We first begin by a claim. Claim: If $(a,b)$ is $n$-good, and $n=xy$ for $(x,y)=1$, then, $(a,b)$ is $x$-good and $y$-good. Proof: To see this, suppose that, $(a,b)$ is $n$-good, but not $x$-good, namely, there exists a pair $(m,k)$ such that, $$ m\not\equiv k \pmod{x} \quad \text{but} \quad P(m)\equiv P(k) \pmod{n}. $$Now, since $(x,y)=1$, the following congruence, $m'\equiv m \pmod{x}$, $m'\equiv 0 \pmod{y}$ is solvable. Similarly, $n'\equiv n \pmod{x}$ and $n'\equiv 0 \pmod{y}$ is solvable. Since, $m' \equiv n' \pmod{y}$, we also have, $P(m')\equiv P(n') \pmod{y}$. On the other hand, we also have, $P(m')\equiv P(n')\pmod{x}$, hence, $P(m')\equiv P(n')\pmod{n}$, while, $m'\not\equiv n' \pmod{n}$, hence, we arrive at a contradiction. As a corollary, if $(a,b)$ is $2010$-good, then, it is also $67$-good. Claim: If $(a,b)$ is $67$-good, then, $67\mid a$ and $67\nmid b$. Proof: This part is the fundamental part. We have the following implication: $$ am^3 + bm \equiv an^3 +bn\pmod{67} \implies m\equiv n \pmod{67}. $$Let us first parse this. $$ am^3 + bm \equiv an^3 +bn\pmod{67} \implies (m-n)(a(m^2+mn+n^2)+b)\equiv 0 \pmod{67}. $$Hence, in order to be able to deduce that $m\equiv n\pmod{67}$, we must have, $a(m^2+mn+n^2)+b \not\equiv 0 \pmod{67}$, for every pair $(m,n)$, for which, $m\not\equiv n \pmod{67}$. We will first show that, if $67\mid b$, then, this quantity can be made to be 0 in modulo $67$, therefore, giving a contradiction. For the purposes, that will be clear soon, it is better to study, $$ 4a(m^2+mn+n^2)+4b = a((2m+n)^2+3n^2)+4b. $$Since, $67 \equiv 1 \pmod{6}$, then, $-3$ is a quadratic residue, hence, there is a $q$ such that, $q^2 \equiv -3 \pmod{p}$. With this $q$, first, fix an $n$, then select $m$ such that, $$ \frac{2m+n}{n} \equiv q \pmod{67}. $$Let us just make sure that, $m\not\equiv n$ is preserved. Indeed, had it been otherwise, we would have had, $q\equiv 3$, contradiction. Hence, for this pair, we have, $$ a((2m+n)^2+3n^2)+4b\equiv a\underbrace{(q^2+3)}_{\equiv 0 \pmod{67}}n^2 + 4b \pmod{67}. $$Hence, if $67\mid b$, we would have a contradiction. First claim is done. Now, we will show that $67 \mid a$. Suppose it is not, and therefore, the number, $\frac{-4b}{3a}$ makes sense in modulo $67$. We will use Legendre's symbol during the proof. First, note that, $(3/67)=(-3/67)(-1/67)=-1$, since, $67\equiv 3\pmod{4}$ and $67 \equiv 1 \pmod{6}$ (could also be deduced by quadratic reciprocity). Next, note that, $$ \left(\frac{-4b/a}{p}\right)=\left(\frac{3}{p}\right)\left(\frac{-4b/3a}{p}\right)\implies \left(\frac{-4b/a}{p}\right) = -\left(\frac{-4b/3a}{p}\right). $$We will now analyze the two cases. $\bullet$ If $\displaystyle\left(\frac{-4b/3a}{p}\right)=1$, meaning, there exists an $n_0$ such that, $n_0^2 \equiv -4b/3a \pmod{67}$, take $n=n_0$, and observe that, $$ a(2m+n_0)^2 + 3an_0^2 + 4b\equiv a(2m+n_0)^2 \pmod{67}. $$Now, observe that, in this case, we can let $m\equiv -n_0/2 \pmod{67}$, and get that, for this pair, $(-n_0/2,n_0)$, we can make the quantity of interest, divisible by $67$, a contradiction. $\bullet$ Now, suppose that, $\displaystyle\left(\frac{-4b/3a}{p}\right)=-1$, namely, $\displaystyle\left(\frac{-4b/a}{p}\right)=1$. Letting $n_0$ such that, $n_0^2 \equiv -4b/a \pmod{67}$, we have, $$ a(2m+n_0)^2 + 3an_0^2 + 4b\equiv a(2m+n_0)^2 - 8b \pmod{67}. $$We will now, give a pair, $(m,n_0)$, such that, this quantity will indeed be divisible by $67$, signaling, the desired contradiction. First, just, verify that, if $m\equiv n_0$, then, we have, $9an_0^2-8b \equiv -44b \pmod{67}$, which can never be zero. Hence, we have guaranteed that, for this pair, $m\not\equiv n_0 \pmod{67}$. Now, the construction. Claim: Under the case being discussed, $8b/a$ is a quadratic residue, modulo $67$. Proof: First, note that, $$ 1=\left(\frac{-4b/a}{p}\right)=\left(\frac{4}{p}\right)\left(\frac{-1}{p}\right)\left(\frac{b/a}{p}\right)\implies \left(\frac{b/a}{p}\right)=-1, $$since, $p\equiv 3\pmod{4}$, $(-1/p)=-1$, and $4$ clearly is a square. Now, we will finish the claim's proof. $$ \left(\frac{8b/a}{p}\right)=\left(\frac{b/a}{p}\right)\left(\frac{4}{p}\right)\underbrace{\left(\frac{2}{p}\right)}_{=-1}\implies \exists q: q^2 \equiv \frac{8b}{a} \pmod{67}, $$since, $2$ is a quadratic residue modulo $p$, if and only if, $p\equiv \pm 1 \pmod{8}$. Hence, we are done. \end{itemize} Therefore, $67 \mid a$. Claim: If $p$ is a prime, such that, $p\mid a$, $p\nmid b$, then, $(a,b)$ is $p^i$-good, for any $i\in\mathbb{N}$. As a corollary, we have, $(a,b)$ is $67^i$ good for any $i\in\mathbb{N}$, proving the result. Proof: Notice that, if, $$ p^i\mid am^3+bm - an^3 - bn \implies p^i\mid (m-n)(\underbrace{a(m^2+mn+n^2)-b}_{\equiv -b \not\equiv 0 \pmod{p}}), $$implies that, $p^i \mid m-n$. Hence, if $p\mid a$, and $p\nmid b$, then, $$ p^i \mid P(m)-P(n) \implies p^i \mid m-n. $$We are done. Remark A similar problem, which, similarly focuses on the injectivity of the map $n\mapsto P(n)$ in a modulo $p$, with certain conditions on $P(n)$, was appeared in Turkish National Mathematical Olympiad in 2000, as fourth problem. That problem's author is also Dr. Tekman. Remark 2 There is a similar problem, in Brazil 96, which also builds upon injectivity of a certain cubic polynomial in a modulo large prime (that I don't remember right now).

a) Take $P(x) = x^3 - 2601x = x(x+51)(x-51)$. Check that $P(x) \equiv x^3$ is bijective modulo $3$ and $17$, as $17 \equiv 2 \pmod 3$. So $P(x)$ is bijective modulo $51$. But for large enough $n$, there are distinct solutions to $P(x) \equiv 0 \pmod{n}$, so $P$ is not very good. b) Let $p=67$ and check with QR law that $-1=\left( \frac{3}{p}\right)= \left( \frac{1/3}{p} \right)$. We are given that $P(x)$ is bijective modulo $p$. Claim: We have $a \equiv 0 \pmod p$. Proof: All equalities in the following are taken modulo $p$. Clearly $b \neq 0$, as $p \equiv 1 \pmod 3$ implies $ax^3$ is not bijective. Suppose for contradiction's sake that $a \neq 0$ as well. We have $P(x) = x(ax^2 + b)$, meaing $-\frac{b}{a}$ is a quadratic non-residue -- else $P(x)=0$ has distinct solutions. But this means $k=-\frac{4b}{3a}$ is a quadratic residue. So, take $y$ so that $y^2=k$ and let $x = -y/2$. I claim $P(x)=P(y)$. Verify that: \[P(y) = ay^3 + by = -ay\left(\frac{4b}{3a}\right) + by = -\frac{1}{3}by,\]\[P(x) = a\left(-\frac{y}{2}\right)^3 - b\left(\frac{y}{2}\right) = \frac{yb}{6} - \frac{yb}{2} = -\frac{1}{3}by.\]Clearly $x \neq y$ because $y \neq 0$. So $P(x)$ fails to be injective, contradiction, and $a=0$, as claimed. $\Box$ Now we finish the problem using the following: Lemma: If a polynomial $P(x) \in \mathbb{Z}[x]$ is bijective modulo $p$ and $P'(x)$ has no roots modulo $p$, then $P(x)$ is bijective modulo any $p^k$.

It's clear that $P'(x) = 3ax^2 + b$ has no roots modulo $p$, as $a \equiv 0 \pmod p$ and $b \not \equiv 0 \pmod p$. We are given bijectivity modulo $p$, so we are done by the lemma. $\blacksquare$ Part (a) is basically telling us that $67 \equiv 1 \pmod{3}$ is crucial to solve part (b). Then, powers of $67$ is the first thing to try for anyone familiar with these kinds of lifting ideas. The main claim $a \equiv 0 \pmod{67}$ is guessable by trying modulo small primes and seeing that the $x^3$ term adds too much noise, or we could see that it must be true if we are to solve it this way. Indeed, if $a \not \equiv 0 \pmod{67}$, then since $3$ is quadratic non-residue we get that $P'(x)$ has a root which destroys our lifting method. The rest is setting $P(x) \equiv P(y)$, canceling $x-y$, and solving the resulting quadratic.

(a) We claim $(a,b)=(1,-51^2)$ is 51-good but isn't very good. First, because $x^3$ is bijective both mod 3 and mod 17, $(1,-51^2)$ is 51-good. On the other hand, $x^3-51^2x = x(x+51)(x-51)$ is zero when $x=0$ and $x=51$, so for any $n\ge 52$ the polynomial $P$ is not injective modulo $n$, so clearly $(1,-51^2)$ is not very good. From part (a) we get the idea that $x^3$ being bijective modulo 17 (i.e. 17 is 2 mod 3) is important. (b) Because 67, a divisor of 2010, is 1 modulo 3, we want to prove that if a polynomial $P(x)=ax^3+bx$ is 67-good then it's also $67^l$-good for any positive integer $l$. First let's see what we can deduce about $a$ and $b$. Assume that $(a,b)$ is 67-good; \begin{align*} P(m)-P(k) = (m-k)(a(m^2+mk+k^2)+b) \tag{1} \end{align*}so we'd like to see when $a(m^2+mk+k^2)+b = \frac14\left(a((2m+k)^2+3k^2)+4b\right)$ is divisible by 67. Suppose that $a$ is not a multiple of 67; then we see \begin{align*}(2m+k)^2+3k^2 \equiv -4ba^{-1} \mod 67. \tag{2}\end{align*}In a similar flavor as APMO 2014/3, the left hand side of (2) covers all residues mod 67, so the only good case is when $m\equiv k \mod 67$. If this happens, we could let $(m',k') = (m+k,-k) = (2m, -m)$ which is no longer the same thing mod 67 except when $m\equiv k \equiv 0 \mod 67$. But this would imply the right hand side of (2) being also divisible by 67, which could only happen when $b$ is divisible by 67. (Un)fortunately this case can also be ruled out: $P(x)=ax^3$ is not bijective mod 67 thanks to the fact that 67 is 1 mod 3. Therefore we conclude that $a$ is divisible by 67 while $b$ is not. Then the term $a(m^2+mk+k^2)+b$ in (1) can never be divisible by 67, so $67^l \mid P(m)-P(k)$ implies $67^l \mid m-k$, and we are done. (Or we can use Hensel's lemma.)

Note that we just need that \[P(x)=ax^3+bx\pmod n\]is a bijection for it to be $n$-good. Using the Chinese Remainder Theorem, it suffices to look at the primes dividing $n$, (a) We shall show that $(1,-51^2)$ is indeed a valid pair. Consider $P(x)=x^3-51^2x$. Taking this modulo $3$, we get that $P(x)\equiv x^3\pmod 3$, so thus as $x^3$ is a bijection on $\mathbb Z/3\mathbb Z$ (as $3\nmid \phi(3)=2$), we have that $P(x)$ is also a bijection modulo 3. Similarly, as $P(x)\equiv x^3\pmod{17}$ and $x^3$ is a bijection (as $3\nmid\phi(17)=16), $$P(x)$ is a bijection, so the pair is 51-good. Now, further note that $P(51)-P(0)=0$, so thus $n\mid P(51)-P(0)$, so $n\mid 51$, implying the existence of solely finitely $n$. Thus $(1,-51^2)$ is one such pair. (b) We shall show that if some $P(x)$ is 67-good, then it is $67^k$-good. Until further stated, work in $\mathbb Z/67\mathbb Z$. First we show $67\mid a$. Suppose otherwise. As $ax^3$ is not a bijection (as $67$ is $1 \pmod {3}$ , so it divides $x^2+xy+y^2$ for some $x,y$ incongruent modulo $67$), we see that $67\nmid b$. It suffices to show $P'(x)=3ax^2+b$ has no roots, and then using Hensel's Lemma. Note for nonzero $x$, as $P(x)\neq 0$, we get $ax^2+b\neq 0$, so $-\frac ba$ is a non quadratic residue. As $3$ is a quadratic nonresidue, we get there exists $x^2=-\frac{4b}{3a}$. However, $P(x)=P\left(-\frac x2\right)$, a contradiction. Thus $a=0$. Now $P(x)\equiv bx\pmod {67}$, so $P'$ has no roots.(as it is just $b \pmod{67}$ )Thus this is completed by Hensel's Lemma.

Notice that $(a,b)$ is $n-$ good if and only if $P(x)=ax^3+bx$ is a bijection mod $p$. (a) We claim that $(a,b)=(1,-51^2)$. Notice that $P(x)=x(x-51)(x+51)$. Notice that for all $n>51$ we have $n|P(51)-P(0)$ but $n\nmid 51$, hence $P$ is not $n-$ good, which implies that $(a,b)$ is not very good. Now we show that $(a,b)$ is $51-$ good. It suffices to show $Q(x)=x^3$ is surjective mod $51$. It is easy to check that $Q$ is surjective mod $3$. Let $g$ be a primitive root mod $17$, hence if $x=g^a$ then $Q(x)=g^3a$, which is clearly a bijection since $f(a)=3a$ is a bijection mod $16$. Now by Chinese remainder theorem we concluded that $Q$ is surjective mod $51$. (b)Notice that if $(a,b)$ is $2010-$ good then it's also $67-$good, we claim that each $(a,b)$ which is $67-$ good is also very good. CLAIM 1. $67\nmid b$. Proof. Otherwise we must have $Q(x)=ax^3$ is injective modulo $67$. However, let $g$ be a primitive root mod $67$, then $$Q(g^{22})=ag^{66}\equiv a\equiv Q(1)\pmod{67}$$hence it is not injective, contradiction. $\blacksquare$ Obivously if $a=0$ then $(a,b)$ is very good. Hence we may assume $a\neq 0$. Hence $a^{-1}P(x)=x^3+a^{-1}bx$ is also a bijection. Hence we may without loss of generality assume $a=1$ CLAIM 2. For all $x$ we have $67\nmid 3x^2+b$ Proof. Notice that $P(x)=x(x^2+b)$. If $-b$ is a quadratic residue mod $67$, then there exists $k$ sucht that $67|k^2+b$, which implies $P(k)=P(0)\pmod{67}$, contradiction. Hence $-b$ is a quadratic nonresidue mod $67$. Suppose on the contrarty that there exists $x$ such that $3x^2\equiv {-b}\pmod{67}$, then $3x^2$ is a quadratic nonresidue mod $67$. However $$\left(\frac{3x^2}{67}\right)=\left(\frac{3}{67}\right)\left(\frac{x^2}{67}\right)=\left(\frac{3}{67}\right)=\left(\frac{67}{3}\right)\cdot (-1)^{\frac{66\cdot 2}{4}}=1$$by law of quadratic reciprocity, contradiction. $\blacksquare$ CLAIM 3. $(a,b)$ is $67^k$-good for every $k\in\mathbb N$. Proof. Indeed we have assume that $(a,b)$ is $67-$good. Moreover $P'(x)=3x^2+b\neq 0\pmod{67}$, hence CLAIM 3 is proved directly via Hensel's lemma. This completes the proof.

Part (a): I claim $(1, -3\cdot 17^2)$ is 51-good and not very good. Consider if $51 | (m^3-k^3) - 3\cdot 17(m-k)$. Taking $\mod 3$, we either have $m-k\equiv 0\mod 3$ or $m^2 + mk + k^2\equiv 0 \mod 3$. Both situations imply $m\equiv k\mod 3$. Taking $\mod 17$, we either have $m-k\equiv 0\mod 17$ or $m^2 + mk + k^2 \equiv 0\mod 17$. For the first case, $m\equiv k\mod17$, so we're done. For the second case, \[m\equiv \frac{-k \pm k\sqrt{-3}}{2}\]However, $k\sqrt{-3}$ is not defined unless $k = 0$, so $m = 0$ as well. Thus, this pair is $51$-good. To show it is not very good, for $n > 51$, just take $m = 17, k = -34$ and we win. Part (b): I claim $67 | a$. FTSOC, assume $67$ does not divide $a$. Then, I claim for each $r$, there exists some $m,k$ such that $m^2 + mk + k^2 \equiv r, m\not\equiv r\mod 67$. We have \[m = \frac{-k \pm \sqrt{4r - 3k^2}}{2}\]I claim there exists $x, y$ such that $4r = 3x^2 + y^2$. This is because we can rewrite this as $4r = (y + x\sqrt{-3})(y - x\sqrt{-3})$, and since $-3$ is a QR $\mod 67$, this always has solutions. Next, taking $k = x$, we get \[m = \frac{-k \pm y}{2}\]This can never be equal to $m$, unless $k = 0, y = 0$. If $k = 0, y = 0$, then $m = 0$, so $r = 0$. However, we can then just take $m = 7, k = 2$. Thus, our claim is proven. However, this means, since $67\not| a$, then there exists some $m,k$ such that $m^2 + mk + k^2 = -\frac{b}{a}$ and $m\neq k$. This means $(a,b)$ isn't 2010-good, so we must have $67 | a$. If $67| b$, then taking $m = \frac{2010}{67}, k = 0$ gives a contradiction. Therefore, $67$ does not divide $b$. Then, I claim $(a,b)$ is $67^{i}$ good, because \[67^{i} | (m-k)(67a'(m^2 + mk + k^2) - b), b\not\equiv 0\mod 67 \Rightarrow 67^{i} | m-k\]Thus, $(a,b)$ must be very-good if it is 2010-good.