kraDracsO wrote: Let $\mathbb{Z}$ be the set of integers. Find all functions $f:\mathbb{Z}\rightarrow\mathbb{Z}$ such that: $$2023f(f(x))+2022x^2=2022f(x)+2023[f(x)]^2+1$$for each integer $x$. For easier writing, let $a=2022$ Let $P(x)$ be the assertion $(a+1)(f(f(x))-f(x)^2-1)=a(f(x)-x^2-1)$ This implies $a+1|RHS$ and so $a+1|f(x)-x^2-1$ $\forall x$ and so $a+1|f(f(x)-f(x)^2-1$ and so $(a+1)^2|LHS$ And easy induction implies $(a+1)^n|LHS $\forall $n$ and so $LHS=0$ and so $RHS=0$ And so $\boxed{f(x)=x^2+1\quad\forall x\in\mathbb Z}$, which indeed fits.

If you consider a function of degree $n$ as its biggest output it's easy to see that only $n=2,1$, it suffices to choose an $x$ big enough to get this. considering the polynomial $ax^2+bx+c$ the original equation $2023af(x) + 2022x^2 = 2022f(x) + 2023f(x)^2+1$ one gets that $$2023(a-1)f(x)^2 + 2023bf(x)+2023c + 2022x^2 = 2022f(x)+1$$so $a =1$ because of matching polynomial degrees. Then $$2023b(x^2+bx+c)+2023c+2022x^2 = 2022(x^2+bx+c)+1$$so $2022 = 2023b + 2022 \rightarrow b = 0$ Then $2023c + 2022x^2 = 2022(x^2+c)+1 \rightarrow c = 1$ The only solution es $$x^2+1$$

@above, apparently you are not searching for non-polynomial solutions, so this is not a complete proof

a_507_bc wrote: @above, apparently you are not searching for non-polynomial solutions, so this is not a complete proof Consider that you somewhere have an x with exponent $x^{1\pm \epsilon}$ or $x^{2\pm \epsilon}$... One can also rearrange the equation to $2022x^2 = H(x) + 1$ It also works to instead consider that the function is $ax^2 + bx + c(x)$ where is $c(x)$ is the "non-polynomial" part of the function (which would get you the constant solution c(x) =1 contradiction)

godfjock wrote: Consider that you somewhere have an x with exponent $x^{1\pm \epsilon}$ or $x^{2\pm \epsilon}$... Uhh ? What if $f(x)=\left\lfloor e^{2943 \sin x}\right\rfloor$ ? godfjock wrote: It also works to instead consider that the function is $ax^2 + bx + c(x)$ where is $c(x)$ is the "non-polynomial" part of the function (which would get you the constant solution c(x) =1 contradiction) Hemm ... Could you just kindly prove this conclusion "(which would get you the constant solution c(x) =1 contradiction)" ????

My solution was pretty much equivalent to pco's. Define $g(x)=f(x)-x^2-1$ and $h(x)=f(x)^2+f(x)-f(f(x))-x^2$. Then $2002h(x)=g(f(x))$ and $2023h(x)=g(x)$, or $2023g(f(x))=2022g(x)$. For any integer $x$, consider sequence $x_0=x$ and $x_n=f(x_{n-1})$ for $n\geq1$. After trivial induction we have $2023^ng(x_n)=2022^ng(x)$, or $2023^n$ divides $g(x)$ for all positive integer $n$ because $2022$ and $2023$ are clearly coprime. Thus $g(x)=0$ and $f(x)=x^2+1$ is a must for all integer $x$. Substitution in the proposed equation yields both sides equal to $2023x^4+6068x^2+4046$, and we are done.

godfjock wrote: If you consider a function of degree $n$ as its biggest output it's easy to see that only $n=2,1$, it suffices to choose an $x$ big enough to get this. considering the polynomial $ax^2+bx+c$ the original equation $2023af(x) + 2022x^2 = 2022f(x) + 2023f(x)^2+1$ one gets that $$2023(a-1)f(x)^2 + 2023bf(x)+2023c + 2022x^2 = 2022f(x)+1$$so $a =1$ because of matching polynomial degrees. Then $$2023b(x^2+bx+c)+2023c+2022x^2 = 2022(x^2+bx+c)+1$$so $2022 = 2023b + 2022 \rightarrow b = 0$ Then $2023c + 2022x^2 = 2022(x^2+c)+1 \rightarrow c = 1$ The only solution es $$x^2+1$$ congrats. 0/7