Subfield: Number Theory


Question: Given an integer $k \geq 2$, set $a_{1}=1$ and, for every integer $n \geq 2$, let $a_{n}$ be the smallest $x>a_{n-1}$ such that:



$$

x=1+\sum_{i=1}^{n-1}\left\lfloor\sqrt[k]{\frac{x}{a_{i}}}\right\rfloor

$$



Prove that every prime occurs in the sequence $a_{1}, a_{2}, \ldots$

Solution:
We prove that the $a_{n}$ are precisely the $k$ th-power-free positive integers, that is, those divisible by the $k$ th power of no prime. The conclusion then follows.



Let $B$ denote the set of all $k$ th-power-free positive integers. We first show that, given a positive integer $c$,



$$

\sum_{b \in B, b \leq c}\left\lfloor\sqrt[k]{\frac{c}{b}}\right\rfloor=c

$$



To this end, notice that every positive integer has a unique representation as a product of an element in $B$ and a $k$ th power. Consequently, the set of all positive integers less than or equal to $c$ splits into



$$

C_{b}=\left\{x: x \in \mathbb{Z}_{>0}, x \leq c, \text { and } x / b \text { is a } k \text { th power }\right\}, \quad b \in B, b \leq c

$$



Clearly, $\left|C_{b}\right|=\lfloor\sqrt[k]{c / b}\rfloor$, whence the desired equality.



Finally, enumerate $B$ according to the natural order: $1=b_{1}<b_{2}<\cdots<b_{n}<\cdots$. We prove by induction on $n$ that $a_{n}=b_{n}$. Clearly, $a_{1}=b_{1}=1$, so let $n \geq 2$ and assume $a_{m}=b_{m}$ for all indices $m<n$. Since $b_{n}>b_{n-1}=a_{n-1}$ and



$$

b_{n}=\sum_{i=1}^{n}\left\lfloor\sqrt[k]{\frac{b_{n}}{b_{i}}}\right\rfloor=\sum_{i=1}^{n-1}\left\lfloor\sqrt[k]{\frac{b_{n}}{b_{i}}}\right\rfloor+1=\sum_{i=1}^{n-1}\left\lfloor\sqrt[k]{\frac{b_{n}}{a_{i}}}\right\rfloor+1

$$



the definition of $a_{n}$ forces $a_{n} \leq b_{n}$. Were $a_{n}<b_{n}$, a contradiction would follow:



$$

a_{n}=\sum_{i=1}^{n-1}\left\lfloor\sqrt[k]{\frac{a_{n}}{b_{i}}}\right\rfloor=\sum_{i=1}^{n-1}\left\lfloor\sqrt[k]{\frac{a_{n}}{a_{i}}}\right\rfloor=a_{n}-1

$$



Consequently, $a_{n}=b_{n}$. This completes the proof.
For every $n=1,2,3, \ldots$, introduce the function



$$

f_{n}(x)=x-1-\sum_{i=1}^{n-1}\left\lfloor\sqrt[k]{\frac{x}{a_{i}}}\right\rfloor

$$



Denote also by $g_{n}(x)$ the number of the indices $i \leq n$ such that $x / a_{i}$ is the $k$ th power of an integer. Then $f_{n}(x+1)-f_{n}(x)=1-g_{n}(x)$ for every integer $x \geq a_{n}$; hence $f_{n}(x)+1 \geq f_{n}(x+1)$. Moreover, $f_{n}\left(a_{n-1}\right)=-1$ (since $f_{n-1}\left(a_{n-1}\right)=0$ ). Now a straightforward induction shows that $f_{n}(x)<0$ for all integers $x \in\left[a_{n-1}, a_{n}\right)$.



Next, if $g_{n}(x)>0$ then $f_{n}(x) \leq f_{n}(x-1)$; this means that such an $x$ cannot equal $a_{n}$. Thus $a_{j} / a_{i}$ is never the $k$ th power of an integer if $j>i$.



Now we are prepared to prove by induction on $n$ that $a_{1}, a_{2}, \ldots, a_{n}$ are exactly all $k$ thpower-free integers in $\left[1, a_{n}\right]$. The base case $n=1$ is trivial.







Assume that all the $k$ th-power-free integers on $\left[1, a_{n}\right]$ are exactly $a_{1}, \ldots, a_{n}$. Let $b$ be the least integer larger than $a_{n}$ such that $g_{n}(b)=0$. We claim that: (1) $b=a_{n+1}$; and (2) $b$ is the least $k$ th-power-free number greater than $a_{n}$.



To prove (1), notice first that all the numbers of the form $a_{j} / a_{i}$ with $1 \leq i<j \leq n$ are not $k$ th powers of rational numbers since $a_{i}$ and $a_{j}$ are $k$ th-power-free. This means that for every integer $x \in\left(a_{n}, b\right)$ there exists exactly one index $i \leq n$ such that $x / a_{i}$ is the $k$ th power of an integer (certainly, $x$ is not $k$ th-power-free). Hence $f_{n+1}(x)=f_{n+1}(x-1)$ for each such $x$, so $f_{n+1}(b-1)=f_{n+1}\left(a_{n}\right)=-1$. Next, since $b / a_{i}$ is not the $k$ th power of an integer, we have $f_{n+1}(b)=f_{n+1}(b-1)+1=0$, thus $b=a_{n+1}$. This establishes (1).



Finally, since all integers in $\left(a_{n}, b\right)$ are not $k$ th-power-free, we are left to prove that $b$ is $k$ th-power-free to establish (2). Otherwise, let $y>1$ be the greatest integer such that $y^{k} \mid b$; then $b / y^{k}$ is $k$ th-power-free and hence $b / y^{k}=a_{i}$ for some $i \leq n$. So $b / a_{i}$ is the $k$ th power of an integer, which contradicts the definition of $b$.



Thus $a_{1}, a_{2}, \ldots$ are exactly all $k$ th-power-free positive integers; consequently all primes are contained in this sequence.