Subfield: Number Theory


Question: Let $n>1$ be a given integer. Prove that infinitely many terms of the sequence $\left(a_{k}\right)_{k \geqslant 1}$, defined by

$$
a_{k}=\left\lfloor\frac{n^{k}}{k}\right\rfloor
$$

are odd. (For a real number $x,\lfloor x\rfloor$ denotes the largest integer not exceeding $x$.)

Solution:
If $n$ is odd, let $k=n^{m}$ for $m=1,2, \ldots$ Then $a_{k}=n^{n^{m}-m}$, which is odd for each $m$.

Henceforth, assume that $n$ is even, say $n=2 t$ for some integer $t \geqslant 1$. Then, for any $m \geqslant 2$, the integer $n^{2^{m}}-2^{m}=2^{m}\left(2^{2^{m}-m} \cdot t^{2^{m}}-1\right)$ has an odd prime divisor $p$, since $2^{m}-m>1$. Then, for $k=p \cdot 2^{m}$, we have

$$
n^{k}=\left(n^{2^{m}}\right)^{p} \equiv\left(2^{m}\right)^{p}=\left(2^{p}\right)^{m} \equiv 2^{m}
$$

where the congruences are taken modulo $p\left(\right.$ recall that $2^{p} \equiv 2(\bmod p)$, by FERMAT's little theorem). Also, from $n^{k}-2^{m}<n^{k}<n^{k}+2^{m}(p-1)$, we see that the fraction $\frac{n^{k}}{k}$ lies strictly between the consecutive integers $\frac{n^{k}-2^{m}}{p \cdot 2^{m}}$ and $\frac{n^{k}+2^{m}(p-1)}{p \cdot 2^{m}}$, which gives

$$
\left\lfloor\frac{n^{k}}{k}\right\rfloor=\frac{n^{k}-2^{m}}{p \cdot 2^{m}}
$$

We finally observe that $\frac{n^{k}-2^{m}}{p \cdot 2^{m}}=\frac{\frac{n^{k}}{2^{m}}-1}{p}$ is an odd integer, since the integer $\frac{n^{k}}{2^{m}}-1$ is odd (recall that $k>m$ ). Note that for different values of $m$, we get different values of $k$, due to the different powers of 2 in the prime factorisation of $k$.
Treat the (trivial) case when $n$ is odd.
Now assume that $n$ is even and $n>2$. Let $p$ be a prime divisor of $n-1$.

Proceed by induction on $i$ to prove that $p^{i+1}$ is a divisor of $n^{p^{i}}-1$ for every $i \geqslant 0$. The case $i=0$ is true by the way in which $p$ is chosen. Suppose the result is true for some $i \geqslant 0$. The factorisation

$$
n^{p^{i+1}}-1=\left(n^{p^{i}}-1\right)\left[n^{p^{i}(p-1)}+n^{p^{i}(p-2)}+\cdots+n^{p^{i}}+1\right]
$$

together with the fact that each of the $p$ terms between the square brackets is congruent to 1 modulo $p$, implies that the result is also true for $i+1$.

Hence $\left\lfloor\frac{n^{p^{i}}}{p^{i}}\right\rfloor=\frac{n^{p^{i}}-1}{p^{i}}$, an odd integer for each $i \geqslant 1$.

Finally, we consider the case $n=2$. We observe that $3 \cdot 4^{i}$ is a divisor of $2^{3 \cdot 4^{i}}-4^{i}$ for every $i \geqslant 1$ : Trivially, $4^{i}$ is a divisor of $2^{3 \cdot 4^{i}}-4^{i}$, since $3 \cdot 4^{i}>2 i$. Furthermore, since $2^{3 \cdot 4^{i}}$ and $4^{i}$ are both congruent to 1 modulo 3 , we have $3 \mid 2^{3 \cdot 4^{i}}-4^{i}$. Hence, $\left\lfloor\frac{2^{3 \cdot 4^{i}}}{3 \cdot 4^{i}}\right\rfloor=\frac{2^{3 \cdot 4^{i}}-4^{i}}{3 \cdot 4^{i}}=\frac{2^{3 \cdot 4^{i}-2 i}-1}{3}$, which is odd for every $i \geqslant 1$.
Treat the (trivial) case when $n$ is odd.
Let $n$ be even, and let $p$ be a prime divisor of $n+1$. Define the sequence $\left(a_{i}\right)_{i \geqslant 1}$ by

$$
a_{i}=\min \left\{a \in \mathbb{Z}_{>0}: 2^{i} \text { divides } a p+1\right\}
$$

Recall that there exists $a$ with $1 \leqslant a<2^{i}$ such that $a p \equiv-1\left(\bmod 2^{i}\right)$, so each $a_{i}$ satisfies $1 \leqslant a_{i}<2^{i}$. This implies that $a_{i} p+1<p \cdot 2^{i}$. Also, $a_{i} \rightarrow \infty$ as $i \rightarrow \infty$, whence there are infinitely many $i$ such that $a_{i}<a_{i+1}$. From now on, we restrict ourselves only to these $i$.

Notice that $p$ is a divisor of $n^{p}+1$, which, in turn, divides $n^{p \cdot 2^{i}}-1$. It follows that $p \cdot 2^{i}$ is a divisor of $n^{p \cdot 2^{i}}-\left(a_{i} p+1\right)$, and we consequently see that the integer $\left\lfloor\frac{n^{p \cdot 2^{i}}}{p \cdot 2^{i}}\right\rfloor=\frac{n^{p \cdot 2^{i}}-\left(a_{i} p+1\right)}{p \cdot 2^{i}}$ is odd, since $2^{i+1}$ divides $n^{p \cdot 2^{i}}$, but not $a_{i} p+1$.