Let $P(x,y)$ denote the given assertion. $P(1,0)$ gives that $f(0) = f(0)(f(1)-1)$, so either $f(0) = 0$ or $f(1) = 2$, if $f(0) = 0$, then $P(x,0)$ gives that $f(x) = 0$ for all $x$, which is indeed a solution. So assume $f(1) = 2$. $P(1,y)$ gives that $f(2y) + f(0) = 2f(y)$, so $f(2) = 2f(1) - f(0) = 4 - f(0)$. But also, from $P(2,0)$, we have that $f(2) = \frac{2}{f(0)} + 1$, equating the two, we get that $f(0)$ is either $1$ or $2$. First suppose $f(0) = 2$, then $P(x,0)$ gives that $2+f(x-1) = 2f(x)$ (which by induction gives $f(n) = 2$ for all $n \in \mathbb{Z}$) and $P(0,x)$ gives $f(2x) + 2 = 2f(x)$ so $f(x-1) = f(2x)$. But $f(x-1) = f(2x) = \frac{f(2x-1)}{2} + 1 = \frac{f(2x-2)}{4} + \frac{3}{2} = \frac{f(x-1) + 2}{2}$, giving that $f(x) = 2$ for all $x$, which is also a solution. Now finally suppose $f(0) = 1$, from $P(x,0)$ we have $f(x) = f(x-1) + 1$ and so by induction $f(n) = n+1$ for all integers $n$. So if $g(x) = f(x) - 1$, then $P(n,x)$ gives that $g(nx) = ng(x)$. Also, since $f(x)$ is between $-2022$ and $2022$ for all $x \in (0,1)$, this means by the above equation that $x - 2023 < g(x) < x+2023$. But $nx - 2023 < g(nx) = ng(x) = g(nx) < nx + 2023$ so we get $x - \frac{2023}{n}< g(x) < x + \frac{2023}{n}$, so taking $n$ sufficiently large, we get that $g(x) = x$ and hence $f(x) = x+1$, which is also a solution. So the only solutions are $f(x) = 0, f(x) = 2$ and $f(x) = x+1$

$P(0,1) : 2f(0) = f(0)f(1)$ so either $f(1) = 2$ or $f(0) = 0$ if $f(0) = 0$ then $P(x,0) : f(x-1) = 0 \implies f(x) = 0$ which clearly is an answer. Now assume $f(1) = 2$. $P(2,0) : f(0) + 2 = f(2)f(0)$ and $P(1,1) : f(2) + f(0) = 4 \implies 4f(0) = f(0)^2 + f(2)f(0) = f(0)^2 + f(0) + 2 \implies f(0)^2 - 3f(0) + 2 = 0 \implies f(0) = 1$ or $2$. if $f(0) = 2$ then $P(1,y) : f(2y) = 2(f(y)-1)$ and $P(x,0) : 2 + f(x-1) = 2f(x) \implies f(x-1) = f(2x)$ but $f(2x) = 1 + \frac{f(2x-1)}{2} = \frac{3}{2} + \frac{f(2x-2)}{4} = \frac{3}{2} + \frac{f(x-1)-1}{2}$ so $f(x-1) = 2$ so $f(x) = 2$ which clearly is an answer. Now assume that $f(0) = 1$. Let $g(x) = f(x) - 1$ Now $P(1,y) : f(2y) + 1 = 2f(y) \implies g(2y) = 2g(y)$ and $P(x,0) : 1 + f(x-1) = f(x) \implies g(x) = g(x-1) + 1$ so for integers $n$ we have $g(2n)= 2f(n)$ and $g(2n) = g(n) + n$ which implies $g(n) = n$ so for integer $n$ we have $f(n) = n+1$. $P(n,y) : f(y(n+1)) + n = f(y)(n+1) \implies g(y(n+1)) = ng(y)$. since $x-2023 < g(x) < x+2023$ for $0 < x < 1$ we have $xn-2023 < ng(x) < xn+2023$ which with increasing $n$ we would have $g(x) = x$ so $f(x) = x+1$. Answers $: f(x) = 0, f(x) = 2, f(x) = x+1$

L567 wrote: Let $P(x,y)$ denote the given assertion. $P(1,0)$ gives that $f(0) = f(0)(f(1)-1)$, so either $f(0) = 0$ or $f(1) = 2$, if $f(0) = 0$, then $P(x,0)$ gives that $f(x) = 0$ for all $x$, which is indeed a solution. So assume $f(1) = 2$. $P(1,y)$ gives that $f(2y) + f(0) = 2f(y)$, so $f(2) = 2f(1) - f(0) = 4 - f(0)$. But also, from $P(2,0)$, we have that $f(2) = \frac{2}{f(0)} + 1$, equating the two, we get that $f(0)$ is either $1$ or $2$. First suppose $f(0) = 2$, then $P(x,0)$ gives that $2+f(x-1) = 2f(x)$ (which by induction gives $f(n) = 2$ for all $n \in \mathbb{Z}$) and $P(0,x)$ gives $f(2x) + 2 = 2f(x)$ so $f(x-1) = f(2x)$. But $f(x-1) = f(2x) = \frac{f(2x-1)}{2} + 1 = \frac{f(2x-2)}{4} + \frac{3}{2} = \frac{f(x-1) + 2}{2}$, giving that $f(x) = 2$ for all $x$, which is also a solution. Now finally suppose $f(0) = 1$, from $P(x,0)$ we have $f(x) = f(x-1) + 1$ and so by induction $f(n) = n+1$ for all integers $n$. So if $g(x) = f(x) - 1$, then $P(n,x)$ gives that $g(nx) = ng(x)$. Also, since $f(x)$ is between $-2022$ and $2022$ for all $x \in (0,1)$, this means by the above equation that $x - 2023 < g(x) < x+2023$. But $nx - 2023 < g(nx) = ng(x) = g(nx) < nx + 2023$ so we get $x - \frac{2023}{n}< g(x) < x + \frac{2023}{n}$, so taking $n$ sufficiently large, we get that $g(x) = x$ and hence $f(x) = x+1$, which is also a solution. So the only solutions are $f(x) = 0, f(x) = 2$ and $f(x) = x+1$ Sorry, I don't see where $nx - 2023 < g(nx) = ng(x) = g(nx) < nx + 2023$ came from, for this to be true wouldn't $nx$ have to be less than 1? I don't see how to prove that, especially as we are taking arbitrarily large values of n