Problem 1378

## Solution

Call the polynomial $p(x)=p_{0}+p_{1} x+p_{2} x^{2}+\ldots+p_{m} x^{m}$. Since $p(x)-n=0$ has -2 and -5 as roots, it must have the factor $(x+2)(x+5)=x^{2}+7 x+10$. So for some $a_{0}, a_{1}, a_{2}, \ldots$ we have:

```
$10 a_{0} \quad+n=p_{0} \in\{0,1,2,3,4,5,6,7,8,9\}$
$10 a_{1}+7 a_{0} \quad=p_{1} \in\{0,1,2,3,4,5,6,7,8,9\}$
$10 a_{2}+7 a_{1}+a_{0} \quad=p_{2} \in\{0,1,2,3,4,5,6,7,8,9\}$
$10 a_{3}+7 a_{2}+a_{1} \quad=p_{3} \in\{0,1,2,3,4,5,6,7,8,9\}$
. .
$10 a_{r+1}+7 a_{r}+a_{r-1}=p_{r+1} \in\{0,1,2,3,4,5,6,7,8,9\}$
. . .
```

Now these equations uniquely determine $\mathrm{a}_{\mathrm{i}}$ and $\mathrm{p}_{\mathrm{i}}$. For $\mathrm{p}_{0}$ must be chosen so that $\mathrm{p}_{0}-\mathrm{n}$ is a multiple of 10 , which fixes $p_{0}$ and $a_{0}$ uniquely. Similarly, given $p_{i}$ and $a_{i}$ for $0 \leq i \leq r$, we have $p_{r+1}=10 a_{r+1}+7 a_{r}+a_{r-1}=$ $7 a_{r}+a_{r-1} \bmod 10$, so $p_{r+1}$ is uniquely determined and hence also $a_{r+1}$. Thus any solution is certainly unique, but it is not clear that the process terminates, so that $\mathrm{p}_{\mathrm{i}}$ and $\mathrm{a}_{\mathrm{i}}$ are zero from some point on.

Evidently the sequence $\mathrm{a}_{\mathrm{i}}$ is bounded. For if $\mathrm{a}_{\mathrm{i}}, \mathrm{a}_{i+1} \leq \mathrm{B} \geq 9$, then $\left|\mathrm{a}_{i+2}\right| \leq 0.7 \mathrm{~B}+0.1 \mathrm{~B}+0.1 \mathrm{~B} \leq \mathrm{B}$. So if we take $B=\max \left(9,\left|a_{0}\right|,\left|a_{1}\right|\right)$, then $\left|a_{i}\right| \leq B$ for all i.

So we can define $\mathrm{L}_{\mathrm{k}}=\min \left(\mathrm{a}_{\mathrm{k}}, \mathrm{a}_{\mathrm{k}+1}, \mathrm{a}_{\mathrm{k}+2}, \ldots\right), \mathrm{U}_{\mathrm{k}}=\max \left(\mathrm{a}_{\mathrm{k}}, \mathrm{a}_{\mathrm{k}+1}, \mathrm{a}_{\mathrm{k}+2}, \ldots\right)$. Obviously, we have $\mathrm{L}_{0} \leq \mathrm{L}_{1} \leq \ldots \leq$ $\mathrm{L}_{\mathrm{k}} \leq \mathrm{L}_{\mathrm{k}+1} \leq \ldots \leq \mathrm{U}_{\mathrm{k}+1} \leq \mathrm{U}_{\mathrm{k}} \leq \ldots \leq \mathrm{U}_{1} \leq \mathrm{U}_{0}$. So $\mathrm{L}_{\mathrm{i}}$ is an increasing integer sequence which is bounded above, so we must have $\mathrm{L}_{\mathrm{i}}=\mathrm{L}$ for all sufficiently large i. Similarly, $\mathrm{U}_{\mathrm{i}}=\mathrm{U}$ for all sufficiently large $\mathrm{i}$, and $\mathrm{L} \leq \mathrm{U}$ (1). But if $\mathrm{a}_{\mathrm{i}}, \mathrm{a}_{\mathrm{i}+1} \geq \mathrm{L}$, then $\mathrm{a}_{\mathrm{i}+2} \leq-0.7 \mathrm{~L}-0.1 \mathrm{~L}+0.9 \leq-0.8 \mathrm{~L}+0.9$. So $\mathrm{U} \leq-0.8 \mathrm{~L}+0.9$ (2). Similarly, if $\mathrm{a}_{\mathrm{i}}, \mathrm{a}_{\mathrm{i}+1} \leq \mathrm{U}$, then $\mathrm{a}_{i+2} \geq-0.8 \mathrm{U}$, so $\mathrm{L} \geq-0.8 \mathrm{U}(3)$.

!

But as the diagram shows the only lattice point satisfying (1), (2), (3) is $(0,0)$, so $a_{i}=0$ for all sufficiently large i, which establishes existence.