Subfield: Number Theory


Question: Let $k$ be a positive integer. Show that if there exists a sequence $a_{0}, a_{1}, \ldots$ of integers satisfying the condition

$$
a_{n}=\frac{a_{n-1}+n^{k}}{n} \quad \text { for all } n \geq 1 \text {, }
$$

then $k-2$ is divisible by 3 .

Solution:
Part $A$. For each positive integer $k$, there exists a polynomial $P_{k}$ of degree $k-1$ with integer coefficients, i. e., $P_{k} \in \mathbb{Z}[x]$, and an integer $q_{k}$ such that the polynomial identity

$$
x P_{k}(x)=x^{k}+P_{k}(x-1)+q_{k}
\tag{I_k}
$$

is satisfied. To prove this, for fixed $k$ we write

$$
P_{k}(x)=b_{k-1} x^{k-1}+\cdots+b_{1} x+b_{0}
$$

and determine the coefficients $b_{k-1}, b_{k-2}, \ldots, b_{0}$ and the number $q_{k}$ successively. Obviously, we have $b_{k-1}=1$. For $m=k-1, k-2, \ldots, 1$, comparing the coefficients of $x^{m}$ in the identity $\left(I_{k}\right)$ results in an expression of $b_{m-1}$ as an integer linear combination of $b_{k-1}, \ldots, b_{m}$, and finally $q_{k}=-P_{k}(-1)$.

Part $B$. Let $k$ be a positive integer, and let $a_{0}, a_{1}, \ldots$ be a sequence of real numbers satisfying the recursion given in the problem. This recursion can be written as

$$
a_{n}-P_{k}(n)=\frac{a_{n-1}-P_{k}(n-1)}{n}-\frac{q_{k}}{n} \quad \text { for all } n \geq 1 \text {, }
$$

which by induction gives

$$
a_{n}-P_{k}(n)=\frac{a_{0}-P_{k}(0)}{n !}-q_{k} \sum_{i=0}^{n-1} \frac{i !}{n !} \quad \text { for all } n \geq 1
$$

Therefore, the numbers $a_{n}$ are integers for all $n \geq 1$ only if

$$
a_{0}=P_{k}(0) \quad \text { and } \quad q_{k}=0 \text {. }
$$

Part $C$. Multiplying the identity $\left(I_{k}\right)$ by $x^{2}+x$ and subtracting the identities $\left(I_{k+1}\right),\left(I_{k+2}\right)$ and $q_{k} x^{2}=q_{k} x^{2}$ therefrom, we obtain

$$
x T_{k}(x)=T_{k}(x-1)+2 x\left(P_{k}(x-1)+q_{k}\right)-\left(q_{k+2}+q_{k+1}+q_{k}\right),
$$

where the polynomials $T_{k} \in \mathbb{Z}[x]$ are defined by $T_{k}(x)=\left(x^{2}+x\right) P_{k}(x)-P_{k+1}(x)-P_{k+2}(x)-q_{k} x$. Thus

$$
x T_{k}(x) \equiv T_{k}(x-1)+q_{k+2}+q_{k+1}+q_{k} \bmod 2, \quad k=1,2, \ldots
$$



Comparing the degrees, we easily see that this is only possible if $T_{k}$ is the zero polynomial modulo 2, and

$$
q_{k+2} \equiv q_{k+1}+q_{k} \bmod 2 \quad \text { for } k=1,2, \ldots
$$

Since $q_{1}=-1$ and $q_{2}=0$, these congruences finish the proof.
Part $A$ and $B$. Let $k$ be a positive integer, and suppose there is a sequence $a_{0}, a_{1}, \ldots$ as required. We prove: There exists a polynomial $P \in \mathbb{Z}[x]$, i. e., with integer coefficients, such that $a_{n}=P(n), n=0,1, \ldots$, and $\quad x P(x)=x^{k}+P(x-1)$.

To prove this, we write $P(x)=b_{k-1} x^{k-1}+\cdots+b_{1} x+b_{0} \quad$ and determine the coefficients $b_{k-1}, b_{k-2}, \ldots, b_{0}$ successively such that

$$
x P(x)-x^{k}-P(x-1)=q,
$$

where $q=q_{k}$ is an integer. Comparing the coefficients of $x^{m}$ results in an expression of $b_{m-1}$ as an integer linear combination of $b_{k-1}, \ldots, b_{m}$.

Defining $c_{n}=a_{n}-P(n)$, we get

$$
\begin{aligned}
P(n)+c_{n} & =\frac{P(n-1)+c_{n-1}+n^{k}}{n}, \quad \text { i. e., } \\
q+n c_{n} & =c_{n-1},
\end{aligned}
$$

hence

$$
c_{n}=\frac{c_{0}}{n !}-q \cdot \frac{0 !+1 !+\cdots+(n-1) !}{n !}
$$

We conclude $\lim _{n \rightarrow \infty} c_{n}=0$, which, using $c_{n} \in \mathbb{Z}$, implies $c_{n}=0$ for sufficiently large $n$. Therefore, we get $q=0$ and $c_{n}=0, n=0,1, \ldots$

Part $C$. Suppose that $q=q_{k}=0$, i. e. $x P(x)=x^{k}+P(x-1)$. To consider this identity for $\operatorname{arguments} x \in \mathbb{F}_{4}$, we write $\mathbb{F}_{4}=\{0,1, \alpha, \alpha+1\}$. Then we get

$$
\begin{aligned}
\alpha P_{k}(\alpha) & =\alpha^{k}+P_{k}(\alpha+1) \quad \text { and } \\
(\alpha+1) P_{k}(\alpha+1) & =(\alpha+1)^{k}+P_{k}(\alpha),
\end{aligned}
$$

hence

$$
\begin{aligned}
P_{k}(\alpha) & =1 \cdot P_{k}(\alpha)=(\alpha+1) \alpha P_{k}(\alpha) \\
& =(\alpha+1) P_{k}(\alpha+1)+(\alpha+1) \alpha^{k} \\
& =P_{k}(\alpha)+(\alpha+1)^{k}+(\alpha+1) \alpha^{k}
\end{aligned}
$$

Now, $(\alpha+1)^{k-1}=\alpha^{k}$ implies $k \equiv 2 \bmod 3$.