Subfield: Algebra


Question: Consider the set

$$
A=\left\{1+\frac{1}{k}: k=1,2,3, \ldots\right\}
$$
For every integer $x \geq 2$, let $f(x)$ denote the minimum integer such that $x$ can be written as the product of $f(x)$ elements of $A$, which are not necessarily different.

Prove that there exist infinitely many pairs $(x, y)$ of integers with $x \geq 2, y \geq 2$, and

$$
f(x y)<f(x)+f(y) .
$$

(Pairs $\left(x_{1}, y_{1}\right)$ and $\left(x_{2}, y_{2}\right)$ are different if $x_{1} \neq x_{2}$ or $\left.y_{1} \neq y_{2}\right)$.

Solution:
Notice that for any positive integer $k$ we have

$$
f\left(2^{k}+1\right) \leq k+1,
$$

because $2^{k}+1=\left(1+\frac{1}{2^{k}}\right) \cdot 2^{k}$ is a representation of $2^{k}+1$ as a product of $k+1$ elements of $A$. We claim that all the pairs $(x, y)$ of the form

$$
x=5, \quad y=\frac{2^{4 k+2}+1}{5}
$$

satisfy the required inequality. Notice that $y$ is an integer for any positive value of $k$, because $2^{4 k+2}+1 \equiv 16^{k} \cdot 4+1 \equiv 5 \equiv 0(\bmod 5)$. Furthermore, $f(x y)=f\left(2^{4 k+2}+1\right) \leq 4 k+3$ (and $f(x)=f\left(2^{2}+1\right) \leq 3$ ) by the above. We now need some lower bounds on the values of $f$. Notice that no element of $A$ exceeds 2 , and therefore the product of at most $k$ elements of $A$ does not exceed $2^{k}$ : it follows that

$$
f(n) \geq\left\lceil\log _{2}(n)\right\rceil \tag{Q2.1}
$$

and in particular that

$$
f(5)=f\left(2^{2}+1\right) \geq\left\lceil\log _{2}(5)\right\rceil=3
$$

We have thus proven $f(x)=f(5)=3$. We want to show $f(x y)<f(x)+f(y)$, and since we know $f(x y) \leq 4 k+3$ and $f(x)=3$ we are reduced to showing $f(y)>4 k$. Since $y>2^{4 k-1}$, from (Q2.1) we already know that $f(y) \geq 4 k$, and hence we just need to exclude that $f(y)=4 k$. Let us assume that we can represent $y$ in the form $a_{1} \cdot \ldots \cdot a_{4 k}$ with every $a_{i}$ in $A$. At least one of the $a_{i}$ is not 2 (otherwise the product would be a power of 2 , while $y$ is odd), and hence it is less than or equal to $3 / 2$. It follows that

$$
a_{1} \cdot \ldots \cdot a_{4 k} \leq 2^{4 k-1} \cdot \frac{3}{2}=15 \cdot \frac{2^{4 k-2}}{5}<\frac{2^{4 k+2}}{5}<y
$$


which contradicts the fact that $a_{1} \cdot \ldots \cdot a_{4 k}$ is a representation of $y$.

Note. Using a similar approach one can also prove that all pairs of the form

$$
\left(3, \frac{2^{2 k+1}+1}{3}\right) \quad \text { and } \quad\left(11, \frac{2^{10 k+5}+1}{11}\right)
$$

satisfy the required inequality.
Notice that for any positive integer $k$ we have

$$
f\left(2^{k}+1\right) \leq k+1,
$$

because $2^{k}+1=\left(1+\frac{1}{2^{k}}\right) \cdot 2^{k}$ is a representation of $2^{k}+1$ as a product of $k+1$ elements of $A$. We claim that all the pairs $(x, y)$ of the form

$$
x=5, \quad y=\frac{2^{4 k+2}+1}{5}
$$

satisfy the required inequality. Notice that $y$ is an integer for any positive value of $k$, because $2^{4 k+2}+1 \equiv 16^{k} \cdot 4+1 \equiv 5 \equiv 0(\bmod 5)$. Furthermore, $f(x y)=f\left(2^{4 k+2}+1\right) \leq 4 k+3$ (and $f(x)=f\left(2^{2}+1\right) \leq 3$ ) by the above. We now need some lower bounds on the values of $f$. Notice that no element of $A$ exceeds 2 , and therefore the product of at most $k$ elements of $A$ does not exceed $2^{k}$ : it follows that

$$
f(n) \geq\left\lceil\log _{2}(n)\right\rceil \tag{Q2.1}
$$

Solution is as in the above section we obtain the lower bound (Q2.1).

Now we claim that all the pairs of the form

$$
x=2^{k}+1, \quad y=4^{k}-2^{k}+1
$$

satisfy the required inequality when $k$ is large enough. To begin with, it is easy to see that

$$
2^{k}+1=\frac{2^{k}+1}{2^{k}} \cdot \underbrace{2 \cdot \ldots \cdot 2}_{k \text { terms }} \quad \text { and } \quad 2^{3 k}+1=\frac{2^{3 k}+1}{2^{3 k}} \cdot \underbrace{2 \cdot \ldots \cdot 2}_{3 k \text { terms }} \text {, }
$$

which shows that $f\left(2^{k}+1\right) \leq k+1$ and $f\left(2^{3 k}+1\right) \leq 3 k+1$. On the other hand, from (Q2.1) we deduce that the previous inequalities are actually equalities, and therefore

$$
f(x)=k+1 \quad \text { and } \quad f(x y)=3 k+1 \text {. }
$$

Therefore, it remain to show that $f(y)>2 k$. Since $y>2^{2 k-1}$ (for $k \geq 1$ ), from (Q2.1) we already know that $f(y) \geq 2 k$, and hence we just need to exclude that $f(y)=2 k$. Let us assume that we can represent $y$ in the form $a_{1} \cdot \ldots \cdot a_{2 k}$. At least one of the factors is not 2 , and hence it is less than or equal to $3 / 2$. Thus when $k$ is large enough it follows that

$$
a_{1} \cdot \ldots \cdot a_{2 k} \leq 2^{2 k-1} \cdot \frac{3}{2}=\frac{3}{4} \cdot 2^{2 k}<2^{2 k}-2^{k}<y
$$

which contradicts the fact that $a_{1} \cdot \ldots \cdot a_{2 k}$ is a representation of $y$.
Let's start by showing that $(x, y)=(7,7)$ satisfies $f(x y)<f(x)+f(y)$. We have $f(7) \geq 4$ since 7 cannot be written as the product of 3 or fewer elements of $A$ : indeed $2^{3}>7$, and any other product of at most three elements of $A$ does not exceed $2^{2} \cdot \frac{3}{2}=6<7$. On the other hand, $f(49) \leq 7$ since $49=2 \cdot 2 \cdot 2 \cdot 2 \cdot 2 \cdot \frac{3}{2} \cdot \frac{49}{48}$.

Suppose by contradiction that there exist only finitely many pairs $(x, y)$ that satisfy $f(x y)<$ $f(x)+f(y)$. This implies that there exists $M$ large enough so that whenever $a>M$ or $b>M$ holds we have $f(a b)=f(a)+f(b)$ (indeed, it is clear that the reverse inequality $f(a b) \leq f(a)+f(b)$ is always satisfied).

Now take any pair $(x, y)$ that satisfies $f(x y)<f(x)+f(y)$ and let $n>M$ be any integer. We obtain

$$
f(n)+f(x y)=f(n x y)=f(n x)+f(y)=f(n)+f(x)+f(y),
$$

which contradicts $f(x y)<f(x)+f(y)$.