The idea it to start studying the fonction in set of integer by proving 1/f(x) is an integer for all integer and after let g(x) =1/f(x) then prove g(x) =kx for all integer then just bounding to get the result f(x) =1/kx

Well, it seems like $f(x)=\frac{1}{kx}$ works for all $k \in \mathbb{Z}^+$

hexapr353 wrote: Well, it seems like $f(x)=\frac{1}{kx}$ works for all $k \in \mathbb{Z}^+$ Yeah I mean prove g(x) = kx. My bad

Sketch: 1)ff(x))=x 2)yf(y)|1 so if y is an integer so 1/f(y) is an integer let g(x)=1/f(x) 3) x-y|g(x)-g(y) 4) xy(g(x)-g(y))|g(x)g(y)(x-y) 5) g(x) = g(1)x for all integer x 6)then bounding to get f(x) = f(1)/x

The only solutions are $f(x) = \frac{1}{cx}$, where $c$ is some positive integer constant. First we show these work. The LHS becomes $xy \left( \frac{1}{cx} - \frac{1}{cy} \right) = \frac{y-x}{c} $ and the RHS becomes $x - y$, so clearly the RHS divided by the LHS is an integer, as desired. Now we show these are the only solutions. Setting $y = x$ gives that $0$ divides $x - f(f(x))$, so $f(f(x)) = x$. So we have\[ xy(f(x) - f(y)) \mid x - y\]Let $P(x,y)$ denote this assertion. $P(f(x), f(y)): f(x) f(y) (x - y) \mid f(x) - f(y)$. Hence\[ f(x) f(y) (x - y) \mid f(x) - f(y) \mid \frac{x-y}{xy},\]so $f(x) f(y) \mid \frac{1}{xy}$ for $x \ne y$. Claim: For all integers $x$, $\frac{1}{f(x)}$ is an integer. Proof: $P(x, f(x))$ gives that $xf(x) \mid 1$ or $f(x) = x$, so $x \mid \frac{1}{f(x)}$ ($\frac{1}{f(x)}$ is an integer) or $f(x) = x$ for each integer $x$. Now, if two distinct integers, were fixed points, say $x$ and $y$, then we have $xy \mid \frac{1}{xy}$, so $xy = 1$, absurd. Hence there is exactly one integer that is a fixed point, and all other integers $x$ satisfy that $\frac{1}{f(x)}$ is an integer. If this integer were $1$, we would be done already because $\frac{1}{f(1)}$ is an integer if $f(1) = 1$. Hence the integer that is a fixed point is not $1$. Suppose it's $k$. Then we have $f(k) f(1) \mid \frac{1}{k}$, so $ kf(1) \mid \frac{1}{k}$, implying that $k^2 \mid \frac{1}{f(1)}$. Now $P(k,1)$ gives $k(k - f(1)) \mid k - 1$, so $k - f(1) \mid k - 1$. Now let $f(1) = \frac{1}{c}$ for some integer $c$. We then have $k - \frac{1}{c} \mid k - 1$, so $kc - 1 \mid kc - c$, implying that $kc - 1 \mid c - 1$, so $k = 1$ or $c = 1$, both of which would imply that $f(1) = 1$, contradiction. $\square$ The claim implies that there is at most one integer fixed point and $P(x, f(x))$ on any integer $x\ne 1$ gives that $xf(x) \mid 1$, so $x \mid \frac{1}{f(x)}$ for all positive integers $x$. Let $g(x) = \frac{1}{f(x)}$, and let $f(1) = \frac 1c$ for some positive integer $c$. We have that $g(1) = c$ and $x\mid \frac{1}{f(x)} = g(x)$ for any positive integer $x$. The equation becomes $xy \left( \frac{1}{g(x)} - \frac{1}{g(y)} \right) \mid x - y$, so $xy(g(x) - g(y)) \mid g(x) g(y) (x - y)$. Let $h(x) = \frac{g(x)}{x}$. Note that $h(x)$ is always an integer for positive integers $x$. Additionally, $h(1) = c$. The equation gives $xy(x h(x) - y h(y) ) \mid xy h(x) h(y) (x - y)$, so\[ x h(x) - yh(y) \mid h(x) h(y) (x - y) \]Let $Q(x,y)$ denote this assertion. Claim: For infinitely many large positive integers $x$, we have $h(x) = c$. Proof: Fix $x$ to be some large enough positive integer satisfying $x - 1 = p > c^2 + c$ is a prime. $Q(x,1): xh(x)-c\mid ch(x)(x-1) = cx h(x) - c h(x)$. Hence $x h(x) - c \mid cx h(x) - c h(x) - c(x h(x) - c) = c^2 - c h(x)$. Thus, $x h(x) - c \mid x (c^2 - c h(x)) + c(x h(x) - c) = c^2(x - 1) $. For size reasons, we have $h(x) < c^2$. This gives that $(p + 1) h(p + 1) - c \mid c^2 p $. Since $p+1 > c^2 + c$, the LHS is greater than $c^2$, and thus cannot divide $c^2$, implying that $p\mid (p+1)h(p + 1) - c$. This means that $h(p + 1) \equiv c\pmod p$. Therefore, if $h(p + 1) \ne c$, $h(p + 1)$ is at least $p + c$, which exceeds, $c^2$, absurd. Hence $h(p + 1) = h(x) = c$, as desired. $\square$ Claim: For all positive rationals $r$, we have $h(r) = c$. Proof: Let $r = \frac mn$, for relatively prime positive integers $m,n$. Consider $Q(r,y)$ where $y$ satisfies $h(y) = c$. We see that $r h(r) - yc \mid c h(r) (r - y)$. Now, multiplying both sides by $n$ gives\[ m h \left( \frac mn \right) - yc \cdot n \mid c \cdot h \left( \frac mn \right) (m - yn) = h\left( \frac mn \right) (cm - yc \cdot n)\]Hence, $ m h \left( \frac mn \right) - yc \cdot n \mid h \left( \frac mn \right) (cm - yc \cdot n) - h \left( \frac mn \right) \left( m h \left( \frac mn \right) - yc \cdot n \right) $. Thus,\[ m h \left( \frac mn \right) - yc \cdot n \mid h \left( \frac mn \right) \left(cm - m h \left( \frac mn \right) \right) \]Choosing $y$ to be sufficiently large implies that the RHS is $0$ (as it does not depend on $y$). Thus, $cm = m h\left( \frac mn \right)$, so $h \left( \frac mn \right) = c$. $\square$ To finish, we get that $h(r) = c$, so $g(r) = cr$, meaning $f(r) = \frac{1}{cr}$, as desired. thx @below does this fix work?

megarnie wrote: The only solutions are $f(x) = \frac{1}{cx}$, where $c$ is some positive integer constant. First we show these work. The LHS becomes $xy \left( \frac{1}{cx} - \frac{1}{cy} \right) = \frac{y-x}{c} $ and the RHS becomes $x - y$, so clearly the RHS divided by the LHS is an integer, as desired. Now we show these are the only solutions. Setting $y = x$ gives that $0$ divides $x - f(f(x))$, so $f(f(x)) = x$. So we have\[ xy(f(x) - f(y)) \mid x - y\]Let $P(x,y)$ denote this assertion. $P(f(x), f(y)): f(x) f(y) (x - y) \mid f(x) - f(y)$. Hence\[ f(x) f(y) (x - y) \mid f(x) - f(y) \mid \frac{x-y}{xy},\]so $f(x) f(y) \mid \frac{1}{xy}$ Claim: $\frac{1}{f(x)}$ is an integer for all positive integers $x$ Proof: Suppose otherwise for some $x\in \mathbb N$. Then let $p$ be a prime with $\nu_p(f(x)) > 0$. We have $f(x)^2 \mid \frac{1}{x^2}$, but this would imply that the $\nu_p$ of $\frac{1}{x^2}$ is greater than $0$, absurd. $\square$ Let $g(x) = \frac{1}{f(x)}$, and let $f(1) = \frac 1c$ for some positive integer $c$. First note that $f(x)^2 \mid \frac{1}{x^2}$ implies that $x^2 \mid g(x)^2$, so $x \mid g(x)$ for any rational $x$. The equation becomes $xy \left( \frac{1}{g(x)} - \frac{1}{g(y)} \right) \mid x - y$, so $xy(g(x) - g(y)) \mid g(x) g(y) (x - y)$. Let $h(x) = \frac{g(x)}{x}$. Note that $h(x)$ is always an integer. Additionally, $h(1) = c$. The equation gives $xy(x h(x) - y h(y) ) \mid xy h(x) h(y) (x - y)$, so\[ x h(x) - yh(y) \mid h(x) h(y) (x - y) \]Let $Q(x,y)$ denote this assertion. Claim: For infinitely many large positive integers $x$, we have $h(x) = c$. Proof: Fix $x$ to be some large enough positive integer satisfying $x - 1 = p > c^2 + c$ is a prime. $Q(x,1): xh(x)-c\mid ch(x)(x-1) = cx h(x) - c h(x)$. Hence $x h(x) - c \mid cx h(x) - c h(x) - c(x h(x) - c) = c^2 - c h(x)$. Thus, $x h(x) - c \mid x (c^2 - c h(x)) + c(x h(x) - c) = c^2(x - 1) $. For size reasons, we have $h(x) < c^2$. This gives that $(p + 1) h(p + 1) - c \mid c^2 p $. Since $p+1 > c^2 + c$, the LHS is greater than $c^2$, and thus cannot divide $c^2$, implying that $p\mid (p+1)h(p + 1) - c$. This means that $h(p + 1) \equiv c\pmod p$. Therefore, if $h(p + 1) \ne c$, $h(p + 1)$ is at least $p + c$, which exceeds, $c^2$, absurd. Hence $h(p + 1) = h(x) = c$, as desired. $\square$ Claim: For all positive rationals $r$, we have $h(r) = c$. Proof: Let $r = \frac mn$, for relatively prime positive integers $m,n$. Consider $Q(r,y)$ where $y$ satisfies $h(y) = c$. We see that $r h(r) - yc \mid c h(r) (r - y)$. Now, multiplying both sides by $n$ gives\[ m h \left( \frac mn \right) - yc \cdot n \mid c \cdot h \left( \frac mn \right) (m - yn) = h\left( \frac mn \right) (cm - yc \cdot n)\]Hence, $ m h \left( \frac mn \right) - yc \cdot n \mid h \left( \frac mn \right) (cm - yc \cdot n) - h \left( \frac mn \right) \left( m h \left( \frac mn \right) - yc \cdot n \right) $. Thus,\[ m h \left( \frac mn \right) - yc \cdot n \mid h \left( \frac mn \right) \left(cm - m h \left( \frac mn \right) \right) \]Choosing $y$ to be sufficiently large implies that the RHS is $0$ (as it does not depend on $y$). Thus, $cm = m h\left( \frac mn \right)$, so $h \left( \frac mn \right) = c$. $\square$ To finish, we get that $h(r) = c$, so $g(r) = cr$, meaning $f(r) = \frac{1}{cr}$, as desired. The same idea as mine but f(x)f(y) | 1/xy if x not equal to y

So we don't have $f(x)^2 \mid \frac{1}{x^2}$

Just to explain more to prove 1/f(x) is an integer we can just make x=f(y) and y is integer so yf(y) |1 or f(y) = y So we can prove that there is no two integer fix point and that's all because we get all integer y yf(y) | 1 and there is at most 1 fix point that we can prove it 1 if f(1) =1 because f(x) =1/x is also a solution