GorgonMathDota wrote: Determine all functions $f: \mathbb{R} \to \mathbb{R}$ such that for any $x,y \in \mathbb{R}$ we have \[ f(f(f(x)) + f(y)) = f(y) - f(x) \] Lol Let \(P(x,y)\) denote the assertion of this functional equation. \(P(x,x)\) shows the existence of a real \(r\) for which \(f(r)=0\). \(P(r,r)\) gives us \[f(f(f(r))+f(r))=0\]or \(f(f(0))=0\). Now, \(P(0,y)\) gives us \(f(f(y))=f(y)-f(0)\). Finally, we can rewrite the functional equation as: \[f(f(x)+f(y)-f(0))=f(y)-f(x)\]Call this \(Q(x,y)\). Comparing \(Q(x,y)\) and \(Q(y,x)\) gives us \(f(x)-f(y)=f(y)-f(x)\) or \(f\) is constant. Checking shows that \(f\equiv0\) only works, and so we conclude that it is the only possible solution.

PARENTHESIS!!!! Define $P[x,y]$ as the equation's assertion. $P[x,\ x] $ will give us \[f(f(f(x))+f(x)) = 0 \]$P \big[x, \ f(f(x)) + f(x) \big]$ will give us \[f(f(f(x))) = - f(x)\]$P\big[x, \ f(f(f(x)))\big] $ will give us \[f(0) = -f(x) - f(f(x)) \ \ \ \ \ \ \ \ \ \ \ \ \ (i) \]$P\big[f(x), x\big]$ will give use \[f(0) = f(x) - f(f(x)) \ \ \ \ \ \ \ \ \ \ \ \ \ (ii) \] From $(i)$ and $(ii)$ we can conclude that $f(x) = 0$ Yeah, I got crazy with all that parenthesis lol

DavyDuf wrote: $P\big[x, \ f(f(f(f(x))))\big] $ will give us \[f(0) = f(x) - f(f(x)) \ \ \ \ \ \ \ \ \ \ \ \ \ (i) \] How?

ZETA_in_olympiad wrote: DavyDuf wrote: $P\big[x, \ f(f(f(f(x))))\big] $ will give us \[f(0) = f(x) - f(f(x)) \ \ \ \ \ \ \ \ \ \ \ \ \ (i) \] How? Typo.... Thank you for letting me know... Anyway, here if you want the parenthesis party Well, $P\big[x, \ f(f(f(f(x))))\big] $, we got \begin{align*} f\Biggl(f\biggl(f\Bigl(x\Bigl)\biggl) + f\biggl(f\Bigl(f\bigl(f\big(f(x)\big)\bigl)\Bigl)\biggl)\Biggl) &= f\biggl(f\Bigl(f\bigl(f\big(f(x)\big)\bigl)\Bigl)\biggl) - f(x) \\ f\bigl(f\big(f(x)\big) - f\big(f(x)\big)\bigl) &= - f\big(f(x)\big) - f(x) \\ f(0) &= - f\big(f(x)\big) - f(x) \end{align*}

DavyDuf wrote: Typo.... Thank you for letting me know... Anyway, here if you want the parenthesis party Well, $P\big[x, \ f(f(f(f(x))))\big] $, we got \begin{align*} f\Biggl(f\biggl(f\Bigl(x\Bigl)\biggl) + f\biggl(f\Bigl(f\bigl(f\big(f(x)\big)\bigl)\Bigl)\biggl)\Biggl) &= f\biggl(f\Bigl(f\bigl(f\big(f(x)\big)\bigl)\Bigl)\biggl) - f(x) \\ f\bigl(f\big(f(x)\big) - f\big(f(x)\big)\bigl) &= - f\big(f(x)\big) - f(x) \\ f(0) &= - f\big(f(x)\big) - f(x) \end{align*} How is $f(f(f(f(f(x)))))=-f(f(x))$? I get $f(f(f(f(f(x)))))=f(f(-f(x))).$

Put P(x,f(y)) to deduce that f(f(x))=c-f(x) (*) so from P(x,y) we have that f(c+f(y)-f(x))=f(y)-f(x) for very x,y reals. Put x=y to deduce that f(c)=0 and in * put x=c to deduce that f(0)=c and in the final put P(0,y) to have f(f(y))=f(y)-c=c-f(y) so f(y)=c for every y and plug in (*) to have c=0 So the only sol is f=0 which works.

ZETA_in_olympiad wrote: DavyDuf wrote: Typo.... Thank you for letting me know... Anyway, here if you want the parenthesis party Well, $P\big[x, \ f(f(f(x)))\big] $, we got \begin{align*} f\Biggl(f\biggl(f\Bigl(x\Bigl)\biggl) + f\biggl(f\Bigl(f\bigl(f\big(x\big)\bigl)\Bigl)\biggl)\Biggl) &= f\biggl(f\Bigl(f\bigl(f\big(x\big)\bigl)\Bigl)\biggl) - f(x) \\ f\bigl(f\big(f(x)\big) - f\big(f(x)\big)\bigl) &= - f\big(f(x)\big) - f(x) \\ f(0) &= - f\big(f(x)\big) - f(x) \end{align*} How is $f(f(f(f(f(x)))))=-f(f(x))$? I get $f(f(f(f(f(x)))))=f(f(-f(x))).$ Again, typo... it should be $P\big[x, \ f(f(f(x)))\big] $, then you have $f(f(f(f(x)))) = - f(f(x))$ Let $f(x) = k, \ \ f(f(f(f(x)))) = f(f(f(k))) = - f(k) = -f(f(x))$

DavyDuf wrote: Let $f(x) = k, \ \ f(f(f(f(x)))) = f(f(f(k))) = - f(k) = -f(f(x))$ No, $f$ is not surjective. But since $f(f(f(x)))=-f(x)$ holds for all $x,$ set $x=f(x)$ so that $f(f(f(f(x))))=-f(f(x)).$

The only solution is $\boxed{f\equiv 0}$, which works. $P(x,x): f(f^2(x) + f(x)) = 0$. $P(x, f^2(x) + f(x)): f^3(x) = -f(x)$. $P(f(x),x): f(0) = f(x) - f(f(x))$. This implies $f(f(0)) = 0$. Thus $-f(0) = f^3(0) = f(0)$, so $f(0) = 0$ and $f(f(x)) = f(x)$ for all $x$. The equation can be rewritten as \[f(f(x) + f(y)) = f(y) - f(x)\] Swapping $x$ and $y$ here and comparing gives that \[f(x) - f(y) = f(y) - f(x) = 0,\]so $f$ is constant, which implies $f\equiv 0$.

DavyDuf wrote: PARENTHESIS!!!! Define $P[x,y]$ as the equation's assertion. $P[x,\ x] $ will give us \[f(f(f(x))+f(x)) = 0 \]$P \big[x, \ f(f(x)) + f(x) \big]$ will give us \[f(f(f(x))) = - f(x)\]$P\big[x, \ f(f(f(x)))\big] $ will give us \[f(0) = -f(x) - f(f(x)) \ \ \ \ \ \ \ \ \ \ \ \ \ (i) \]$P\big[f(x), x\big]$ will give use \[f(0) = f(x) - f(f(x)) \ \ \ \ \ \ \ \ \ \ \ \ \ (ii) \] From $(i)$ and $(ii)$ we can conclude that $f(x) = 0$ Yeah, I got crazy with all that parenthesis lol I‘m the same~

Let $P(x,y)$ be the assertion. We claim $f(x)=0$, which clearly works, is the only solution. $P(0,0) \rightarrow f(f(f(0))+f(0))=0 \rightarrow \exists n \in \mathbb{R} \text{ with } f(n)=0.$ $P(n,n) \rightarrow f(f(0))=0$. $P(0,n) \rightarrow f(0)=-f(0) \rightarrow f(0)=0$. $P(0,x) \rightarrow f(f(x))=f(x)$. $P(x,0) \rightarrow f(f(f(x)))=-f(x) \rightarrow f(f(x))=-f(x)$. Thus, $f(x)=-f(x)$, so $f(x)=0$ is the only solution.

Denote $P(x,y)$ be the assertion. Let $f^n(x) =f(f^{n-1}(x)) \quad \forall n \in \mathbb{N}\ge 2 $ with $f^1(x)=f(x)$. For $P(x, x)$, we have $f(f^2(x)+f(x))=0$ For $P(x, f^2(x)+f(x))$, we have $f^3(x)=-f(x) \quad (i)$ For $P(f(x), x) $, we have $f(0)=f(x)-f^2(x) \quad (ii) $ For $P(x,0) $, we have $f^2(x)=-f^2(x) \rightarrow f^2(x)=0$ Therefore, by $(i) $ $f(0)=-f(x)$ Moreover, by $(ii) $ $f(0)=f(x)$ From here, we can conclude that the only solution is $\boxed {f(x) =0 \quad \forall x\in \mathbb{R}} $.

firstly let P(x,y) be the assertion $P(x,x) \rightarrow f(f(f(x))+f(x)) = 0\forall x \in \mathbb{R}$ we pick u so that $f(f(f(u))+f(u))=0 $ $P(x,u) = f(f(f(x))) =-f(x)$ $P(u,y) = f(f(0)+f(y)) = f(y)$ $P(u,u) = f(f(0)) = 0$ $P(0,u) = f(f(f(0))) = -f(0)$ $P(0,u) =2f(0) = 0$ $f(0) = 0$ $P(0,y) = f(f(y)) = f(y)$ $P(x,u) = f(f(f(x)) = -f(x)$ $f(f(f(y)) = f(f(y)) = f(y)$ $f(x) = -f(x)$ so $2f(x) = 0$

GorgonMathDota wrote: Determine all functions $f: \mathbb{R} \to \mathbb{R}$ such that for any $x,y \in \mathbb{R}$ we have \[ f(f(f(x)) + f(y)) = f(y) - f(x) \] $\color{blue}\boxed{\textbf{Answer:}f\equiv\textbf{0}}$ $\color{blue}\boxed{\textbf{Proof:}}$ $\color{blue}\rule{24cm}{0.3pt}$ $$f(f(f(x))+f(y))=f(y)-f(x)...(\alpha)$$In $(\alpha) x=y:$ $$\Rightarrow f(f(f(x))+f(x))=0$$$$\Rightarrow \exists\text{ }a / f(a)=0$$In $(\alpha) x=y=a:$ $$\Rightarrow f(f(0))=0$$In $(\alpha) x=0:$ $$\Rightarrow f(f(y))=f(y)-f(0)...(I)$$$(I)$ in $(\alpha):$ $$\Rightarrow f(f(x)+f(y)-f(0))=f(y)-f(x)...(\beta)$$In $(\beta) x=a:$ $$\Rightarrow f(f(y)-f(0))=f(y)$$$$\Rightarrow f(f(x)-f(0))=f(x)...(II)$$In $(\beta) y=a:$ $$\Rightarrow f(f(x)-f(0))=-f(x)...(III)$$By $(II)$ and $(III):$ $$\Rightarrow f(x)=-f(x)$$$$\Rightarrow \boxed{f(x)\equiv 0}_\blacksquare$$$\color{blue}\rule{24cm}{0.3pt}$

We claim that the answer is \[f \equiv 0\]It is easy to see that these solutions work. Now, we shall show that they are the only ones. Note that, $P(x,x)$ gives \[f(f(f(x))+f(x))=f(x)-f(x)=0\]Now, consider $\alpha$ such that $f(\alpha)=0$. Then, $P(x,\alpha)$ gives \[-f(f(x))=f(f(f(f(x))))=f(-f(x))\]Then, $P(x,-f(x))$ gives us that \[f(0)=f(f(f(x))-f(f(x)))=f(f(f(x))+f(-f(x)))=f(-f(x))-f(x)=-f(f(x))-f(x)\]Thus, for all $x\in \mathbb{R}$, we have $f(f(x))=-f(x)-f(0)$. Now, $P(x,0)$ gives us , \[-(-f(x)-f(0))=-f(f(x))=f(-f(x))=f(f(f(x))+f(0))=f(0)-f(x)\]This gives us that for all $x\in \mathbb{R}$, \[f(x)+f(0)=f(0)-f(x) \implies 2f(x)=0\]Thus $f(x)=0$ for all $x\in \mathbb{R}$ which implies that $f\equiv 0$ is the only solution as claimed.

Let $P(x,y)$ denote the assertion. $P(x,x): f(f(f(x))+f(x))=f(x)-f(x)=0$ so there exists $a$ such that $f(a)=0$ $P(a,a): f(f(0))=0$ $P(0,f(0)): f(0)=-f(0) \Rightarrow f(0)=0$ $P(0,x): f(f(x))=f(x)\quad (1)$ $P(f(x),0):f(f(f(f(x))))=-f(x) \stackrel{(1)} {<=>} f(f(x))=-f(x) \quad (2) $ So by (1) and (2) we get: $f(x)=-f(x) \Rightarrow \boxed{f(x)=0} \forall x \in \mathbb{R}$ (Big thanks to @ItsBesi for correcting me in the first part)

JanHaj wrote: Let $P(x,y)$ denote the assertion. $P(0,0): f(f(f(0))+f(0))=0$ so there exists $a$ such that $f(a)=0$ $P(a,a): f(f(0))=0$ $P(0,f(0)): f(0)=-f(0) \Rightarrow f(0)=0$ $P(0,x): f(f(x))=f(x)\quad (1)$ $P(f(x),0):f(f(f(f(x))))=-f(x) \stackrel{(1)} {<=>} f(f(x))=-f(x) \quad (2) $ So by (1) and (2) we get: $f(x)=-f(x) \Rightarrow \boxed{f(x)=0} \forall x \in \mathbb{R}$ I think you have made a mistake $ f(f(f(0))+f(0))=0$ doesn't imply that there exists $a$ such that $f(a)=0$ But what you can do is take $P(x,x) \implies f(f(x))+f(x))=f(x)-f(x)=0 \implies f(f(f(x))+f(x))=0$ So we get that $0$ must be in the range of $f$ Now we can say that there exists $a$ such that $f(a)=0$ And the rest is the same Edit: It's fixed now

Quite easy FE

here is my solution if i made a mistake reply this

Nice handwriting

Let $P(x, y)$ denote the given assertion. $P(x, x) \implies \exists a \text{ such that } f(a) = 0$. $P(x, a) \implies f^3(x) = -f(x)$. In particular, $P(a, a) \implies f^2(0) = 0$. Now, $P(0, x) \implies f^2(x) = f(x) - f(0)$. So $f^3(x) = f^2(f(x)) = f(f(x)) - f(0) = f(x) - 2f(0)$. But $f^3(x) = -f(x)$. So $-f(x) = f(x) - 2f(0) \implies f(x) = f(0)$ or $f$ is constant. But the only constant solution which works is $f(x) = 0 \forall x$, so this is our only solution. $\square$

Let $P(x, y)$ denote the assertion. From $P(x, x)$, we have $f( f(f(x)) + f(x) ) = 0$, call $f(f(x)) + f(x) = \alpha$. From $P(\alpha, \alpha)$, we have $f(f(0)) = 0$. $P(0, y)$ gives $f(f(y)) = f(y) - f(0)$, $P(x, 0)$ gives $f(f(x)) = f(0) - f(x) \implies f(x) = f(0)$, and since $f$ is constant $f \equiv 0$ by substituting this into the FE.

alexanderhamilton124 wrote: Nice handwriting Tysm