Pre-IMO 2017 Mock Exam

Suppose that $$a_2{b}_2{n}^2+a_1{b}_{1}n+a_0{b}_0\mid (a_2^{2017n}+b_2)n^2+(a_1^{2017n}+b_1)n+(a_0^{2017n}+b_0)(1)$$ 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 $$a_2{b}_2{n}_0^2+a_1{b}_1{n}_0+1.(2)$$ 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 $n\equiv n_0(\mathrm{mod}p)$. Then $p$ divides the left-hand side of (1). This gives $$p\mid (a_2^{2017n}+b_2)n_0^2+(a_1^{2017n}+b_1)n_0+2.(3)$$ 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 $$p\mid (a_2^{2017n}-1)n_0+(a_1^{2017n}-1)(4)$$ for any $n\equiv n_0(\mathrm{mod}p)$. By taking $n\equiv 1(\mathrm{mod}p-1)$ and $n\equiv 2(\mathrm{mod}p-1)$ 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 $n_0\equiv -1(\mathrm{mod}p)$.

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 $n_0\equiv -1(\mathrm{mod}p)$. 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 $n_0\equiv -1(\mathrm{mod}p)$. 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 $a_0=a_1=a_2=b_0=b_1=b_2=1$. $\square$