Problem 1881

Suppose that
$$a_2 b_2 n^2 + a_1 b_1 n + a_0 b_0 \mid (a_2^{2017n} + b_2)n^2 + (a_1^{2017n} + b_1)n + (a_0^{2017n} + b_0) \quad (1)$$
We first claim that $a_0 = b_0 = 1$. Let $n$ be a large multiple of $a_0 b_0$. Since $a_0 b_0$ divides the left-hand side of (1), it also divides the right-hand side and hence $a_0 b_0 \mid a_0^{2017n} + b_0$. This implies $a_0 \mid b_0$ and $b_0 \mid a_0^{2017n}$. This shows $a_0$, $b_0$ have the same prime divisors. Thus, we can take a sufficiently large $n$ such that $a_0 b_0 \mid a_0^{2017n}$. This gives $a_0 b_0 \mid b_0$. The only possibility is $a_0 = 1$, which implies $b_0 = 1$.

Let $M$ be a sufficiently large integer and let $n_0$ be the product of all primes less than $M$. Choose any prime divisor $p$ of
$$a_2 b_2 n_0^2 + a_1 b_1 n_0 + 1. \quad (2)$$
Clearly, $(p, n_0) = 1$. Thus, $p \ge M$. In particular, we have $(p, a_1 a_2) = 1$ as $M$ is large.

Consider any $n$ such that $n \equiv n_0 \pmod{p}$. Then $p$ divides the left-hand side of (1). This gives
$$p \mid (a_2^{2017n} + b_2)n_0^2 + (a_1^{2017n} + b_1)n_0 + 2. \quad (3)$$
We choose $n$ such that $p-1 \mid n$. Such an $n$ exists by the Chinese remainder theorem. By Fermat's little theorem, we obtain $p \mid (1+b_2)n_0^2 + (1+b_1)n_0 + 2$. Taking the difference with (3), as $(p, n_0) = 1$, we get
$$p \mid (a_2^{2017n} - 1)n_0 + (a_1^{2017n} - 1) \quad (4)$$
for any $n \equiv n_0 \pmod{p}$. By taking $n \equiv 1 \pmod{p-1}$ and $n \equiv 2 \pmod{p-1}$ respectively, we have
$$p \mid (a_2^{2017 \times 2} - 1)(a_1^{2017} - 1) - (a_1^{2017 \times 2} - 1)(a_2^{2017} - 1) = (a_1^{2017} - 1)(a_2^{2017} - 1)(a_2^{2017} - a_1^{2017}).$$
Since $p \ge M$ is sufficiently large, the right-hand side must be $0$. If one of $a_1$, $a_2$ is $1$, then (4) implies both are equal to $1$. If $a_1 = a_2$, then (4) implies $a_1 = a_2 = 1$ or $n_0 \equiv -1 \pmod{p}$.

We first consider the case $a_1 = a_2 = 1$. In that case, (1) becomes $b_2 n^2 + b_1 n + 1 \mid (1+b_2)n^2 + (1+b_1)n + 2$ so that $b_2 n^2 + b_1 n + 1 \mid n^2 + n + 1$. Note that $b_2 n^2 + b_1 n + 1 \ge n^2 + n + 1$. Thus, equality must hold and hence $b_1 = b_2 = 1$. One easily checks that $a_0 = a_1 = a_2 = b_0 = b_1 = b_2 = 1$ is a solution.

Next, it remains to consider the case $n_0 \equiv -1 \pmod{p}$. As $p$ divides (2), this yields $p \mid a_2 b_2 - a_1 b_1 + 1$. Since $p$ is large, we must have $a_2 b_2 - a_1 b_1 + 1 = 0$. Then (2) becomes $(a_1 b_1 - 1)n_0^2 + a_1 b_1 n_0 + 1 = (n_0 + 1)((a_1 b_1 - 1)n_0 + 1)$. Therefore, instead of choosing any prime $p$ dividing (2) at the beginning, we choose such a prime $p$ dividing $(a_1 b_1 - 1)n_0 + 1$. Using the same argument, we obtain either the same solution or $n_0 \equiv -1 \pmod{p}$. In the latter case, we find that $p \mid -(a_1 b_1 - 1) + 1$. Again, this forces $a_1 b_1 = 2$ as $p$ is large. Thus, $(a_1, b_1) = (1, 2)$ or $(2, 1)$. Also, $a_2 b_2 = a_1 b_1 - 1 = 1$ so that $a_2 = b_2 = 1$.

Now, by considering $n = 3$ in (1), we have $16 \mid 2(3)^2 + (a_1^{2017n} + b_1)(3) + 2$. As $a_1$, $b_1$ have different parities, the right-hand side is odd. This is impossible.

Therefore, the only solution is $a_0 = a_1 = a_2 = b_0 = b_1 = b_2 = 1$. $\square$