Let $x=0$,$f(y)=f(f(y))(*)$,for all $y\in R$. Let $x=1$,$f(y+1)=f(1)+f(f(y))=f(1)+f(y)$ for all $y\in R$. Let $y=0$,$f(0)=0,f(x^4)=x^3f(x),f(x^4+y)=f(x^4)+f(y)$ Let $a=x^4\geq 0,f(a+y)=f(a)+f(y)$ for all $a>0,y\in R$ Let $y=a$,$f(2a)=2f(a)$,by induction,for all $n\in N,f(na)=nf(a)$ If there exist $x\neq y$,$f(x)=f(y)$,Suppose $x>y,a=x-y>0$. $f(x)=f(a)+f(y),f(a)=0$ So $f(na)=0$ for all $n\in N$, a contradiction with condition (i) So for all $x\neq y$,$f(x)\neq f(y)$. By (*),$f(y)=y$.for all $y\in R$. So $f(x)=x$.

We don't need a to be >0. Because $ ( - x)^3.f( - x) = f(x^4) = x^3.f(x) \rightarrow f(x) = - f( - x)$ So for a<0: $ f( - a) + f( - x) = f( - a - x) \rightarrow - f(a) - f(x) = - f(a + x) \rightarrow f(a) + f(x) = f(a + x)$

Setting $x=0$ shows that $f(f(y)) = f(y)$, so $f(x^4 + y) = x^3 f(x) + f(y)$ for all real $x,y$. If $f(y) > y$ for any $y$, setting $x = (f(y) - y)^{\frac{1}{4}}$ gives $f(f(y)) = f(x^4 + y) = x^3 f(x) + f(f(y))$, whence $x^3 f(x) = 0$, or $f(x) = 0$. Note that $f(x^4 + y) = x^3 f(x) + f(y) = f(y)$ for all $y$, whence $f(kx^4 + x) = f(x) = 0$ for all $k$. Since $f(y) > y$, $x > 0$, so there are infinitely many $z$ ($z = kx^4 + x$, $k \in \mathbb{Z}^+$) for which $f(z) = 0$, which is a contradiction. Hence, $f(y) \leq y$ for all real $y$. Setting $x=1$ gives $f(y+1) = f(y) + f(1)$. Setting $y=0$ in this equation gives $f(0) = 0$. Setting $y=0$ in the original equation gives $f(x^4) = x^3 f(x)$, and setting $y = -x^4$ gives $0 = f(0) = x^3 f(x) + f(-x^4)$. Hence, $f(-x^4) = -f(x^4)$ for all $x$, whence $f(x) = -f(-x)$ for all real $x$. Since $f(x) \leq x$ and $-f(x) = f(-x) \leq -x$ for all real $x$, we find that $f(x) = x$ for all real $x$.

Let $P(x,y)$ denote the given relation from $(ii)$. Setting $P(0,x)$ gives $f(x) = f(f(x))$ for every $x$ in $R$. Setting $P(-1,-1)$ gives us $f(0) = -f(-1) + f(f(-1)) = -f(-1) + f(-1) = 0$. Setting $P(x,0)$ gives $f(x^{4}) = x^{3}f(x)$. Now $P(-x,0) = -x^{3}f(-x)$, so $f$ is odd. The next part is to use $(i)$ to prove that $f(s) = 0$ then $s = 0$. Notice that if there is a number $s$ such that $f(s) = 0$ different from $0$, there is also a positive such number because $f$ is odd. So assume that there is a positive number $s$ such that $f(s) = 0$. Setting $P(s^{\frac{1}{4}}, s)$ gives $f(2s) = f(s) + f(f(s)) = 0$ so $f(2s) = 0$. Now by induction we show that $f(ns) = 0$ for every natural number $n$. But this is easy, just by setting $P((ns)^{\frac{1}{4}}, s)$ gives $f((n+1)s) = f(ns + s) = f(ns) + f(s) = 0$. This violates $(i)$ so if $f(s) = 0$, then $s = 0$. Setting $P(x, f(x^{4}) - x^{4})$ and using $f(f(x)) = f(x)$ we get $f(f(x^{4}) - x^{4}) = 0$ and so $f(x^{4}) = x^{4}$, but $f$ is odd so $f(x) = x$ is the only solution.

steppewolf wrote: Let $P(x,y)$ denote the given relation from $(ii)$. Setting $P(0,x)$ gives $f(x) = f(f(x))$ for every $x$ in $R$. Setting $P(-1,-1)$ gives us $f(0) = -f(-1) + f(f(-1)) = -f(-1) + f(-1) = 0$. Setting $P(x,0)$ gives $f(x^{4}) = x^{3}f(x)$. Now $P(-x,0) = -x^{3}f(-x)$, so $f$ is odd. The next part is to use $(i)$ to prove that $f(s) = 0$ then $s = 0$. Notice that if there is a number $s$ such that $f(s) = 0$ different from $0$, there is also a positive such number because $f$ is odd. So assume that there is a positive number $s$ such that $f(s) = 0$. Setting $P(s^{\frac{1}{4}}, s)$ gives $f(2s) = f(s) + f(f(s)) = 0$ so $f(2s) = 0$. Now by induction we show that $f(ns) = 0$ for every natural number $n$. But this is easy, just by setting $P((ns)^{\frac{1}{4}}, s)$ gives $f((n+1)s) = f(ns + s) = f(ns) + f(s) = 0$. This violates $(i)$ so if $f(s) = 0$, then $s = 0$. Setting $P(x, f(x^{4}) - x^{4})$ and using $f(f(x)) = x$ we get $f(f(x^{4}) - x^{4}) = 0$ and so $f(x^{4}) = x^{4}$, but $f$ is odd so $f(x) = x$ is the only solution. f(x) = -x is odd too, but you can easily rule that case out by plugging in some values.

My solution: $P(0,y)$ gives $f(y) = f^2(y)$. $P(x,1)$ and $P(1,x^4)$ give $f(x^4) = x^3f(x)$. In particular, we also have $f(0)=0$. So we have for all $x \ge 0$, $f(x+y) = f(x) + f(y)$. Plugging in $y=-x$ gives $f$ is odd. So $f$ satisfies Cauchy's equation. Suppose for the sake of contradiction that $f$ is not injective. Let $f(a) = f(b)$ for unequal $a,b$. We have $f(a-b)=0$ from Cauchy's equation. Then by induction using Cauchy's equation, we have $f(n(a-b)) = 0$ for every $n \in \mathbb{N}$. This contradicts the assumption, hence $f$ is injective. So we must have, using the first relation we found, that $f(y) = y$ $\forall \ y \in \mathbb{R}$.

WizardMath wrote: My solution: $P(0,y)$ gives $f(y) = f^2(y)$. $P(x,1)$ and $P(1,x^4)$ give $f(x^4) = x^3f(x)$. In particular, we also have $f(0)=0$. So we have for all $x \ge 0$, $f(x+y) = f(x) + f(y)$. Plugging in $y=-x$ gives $f$ is odd. So $f$ satisfies Cauchy's equation. Suppose for the sake of contradiction that $f$ is not injective. Let $f(a) = f(b)$ for unequal $a,b$. We have $f(a-b)=0$ from Cauchy's equation. Then by induction using Cauchy's equation, we have $f(n(a-b)) = 0$ for every $n \in \mathbb{N}$. This contradicts the assumption, hence $f$ is injective. So we must have, using the first relation we found, that $f(y) = y$ $\forall \ y \in \mathbb{R}$. I did the exact same thing !

Solution.Let $P(x,y)$ be the assertion for $f(x^4+y)=x^3f(x)+f(f(y))$. $$P(0, x)\implies f(x)=f(f(x))\qquad (1)$$$$P(1, x)\implies f(x+1)=f(x)+f(1)\qquad (2)$$Lemma 1. $f$ satisfies Cauchy's equation. Proof. In $(2)$ letting $x=0\implies f(0)=0$. $P(x,0)\implies f(x^4)=x^3f(x)$. Rewriting the original equation gives that $f(x+y)=f(x)+f(y)\forall x\geq 0$. Since $f(0)=0$, plugging $y=-x$ $\implies -f(x)=f(-x)\implies f$ is odd$\implies f$ satisfies Cauchy's equation.$\square$ Lemma 2. If we prove that $f$ is injective we are done. Proof. By $(1)$ if $f$ is injective $\implies f(x)=x\forall x\in \mathbb{R}$. Thus if $f$ is injective we're done.$\square$ Lemma 3. $f$ is injective. Proof.FTSOC suppose $f$ is not injective, so we have that $f(a)=f(b)$, and $a\neq b$. We have that $f(x+y)=f(x)+f(y)$, plugging $x=a$, $y=-b \implies 0=f(a)-f(b)=f(a)+f(-b)=f(a-b)$. Now we have $f(a-b)=0$, by induction $\implies f(k(a-b))=0\forall k\in \mathbb{N}\implies a=b$.$\square\blacksquare$

Let $P(x,y)$ be the given assertion. $P(0,x)\Rightarrow f(f(x))=f(x)$ $P(-1,-1)\Rightarrow f(0)=0$ $P(-x,y)-P(x,y)\Rightarrow x^3f(x)=-x^3f(-x)\Rightarrow f(x)=-f(-x)$ for all $x\ne0$, but it's true for $x=0$ also. Assume $f(k)=0$ for some $k\ne0$. WLOG $k>0$ (by oddness). $P(k,k)\Rightarrow f(k^4+k)=0$ Consider the sequence $k_n$ defined by $k_1=k$ and $k_{n+1}=k_n^4+k_n$. By $P(k_n,k_n)$ we have by induction that all $k_n$ are zeroes of $f$. Also, $k_n>0$ implies $k_{n+1}=k_n^4+k_n>k_n>0$ and so all $k_n$ are distinct. This means that $f$ has infinitely many zeroes which is a contradiction. So $f(k)=0\Leftrightarrow k=0$. $P(x,0)\Rightarrow f(x^4)=x^3f(x)$ $P\left(\sqrt[4]x,y\right)\Rightarrow f(x+y)=f(x)+f(y)\forall x\ge0$. Set $x\mapsto x+y$ and $y=-y$ for the assertion $f(x)+f(y)=f(x+y)$ if $x+y\ge0$. But $f(-x)+f(-y)=f(-x-y)$, so $f$ is additive in general. If $f(a)=f(b)$ for some $a,b$. Setting $x=a-b$ and $y=b$ in this gives $f(a-b)=0$, and so $a-b=0$, implying injectivity. Then $f(f(x))=f(x)\Rightarrow\boxed{f(x)=x}$ which fits.

$P(0,y) : f(y) = f(f(y))$. $P(1,0) : f(1) = f(1) + f(f(0)) \implies f(f(0)) = f(0) = 0$. $P(x,0) , P(-x,0) \implies x^3f(x) = -x^3f(-x) \implies f$ is odd. $f(x^4+y) = x^3f(x) + f(f(y)) \implies f(x^4+y) = f(x^4) + f(y) \implies f$ is a Cauchy function. Assume $n \neq 0$ such that $f(n) = 0$ Now note that we have $f(2n) = 2f(0) = 0$ and with induction we have $f(an) = 0$ for infinite $a$ which gives contradiction so $n$ doesn't exist. Assume $a,b$ such that $f(a) = f(b)$ Now we have $f(x^4+a) = f(x^4) + f(a) = f(x^4) + f(b) = f(x^4+b)$. Now assume that $a \neq b$ then we have $f(0) = f(b-a)$ so $b-a = 0 \implies a = b$ so contradiction so $f$ is injective. Note that we had $f(ax) = af(x) \implies f(f(ax)) = f(af(x)) \implies f(ax) = f(af(x)) \implies ax = af(x) \implies f(x) = x$. Answers $: f(x) = x$.

Let $P(x,y)$ be the assertion in (ii). $P(0,y)\implies f(y)=f(f(y)).$ $P(1,0)\implies f(f(0))=f(0)\implies f(0)=0.$ Comparing $P(x,0)$ and $P(-x,0)$ we get $f$ is odd. If $f(1)=0,$ then by induction $f(n)=0~\forall n\in \mathbb{N},$ a contradiction with (i). Thus $f(1)\neq 0.$ Suppose $\exists f(z)=0: z>0$ and $z \neq 1.$ $P(z,0)\implies f(z^4)=z^3f(z)=0,$ which contradicts (i). Thus $f(x)=0\iff x=0.$ $P(\sqrt[4]{x},y)\implies f$ is additive. Suppose $\exists u,v: f(u)=f(b)$ for some real $u,v.$ $P(u-v,v)\implies f(u-v)=0\implies u=v,$ thus $f$ is injective. $P(0,x)\implies \boxed{f(x)=x}~~\forall x,$ which obviously works.

shobber wrote: Let ${\bf R}$ denote the set of all real numbers. Find all functions $f$ from ${\bf R}$ to ${\bf R}$ satisfying: (i) there are only finitely many $s$ in ${\bf R}$ such that $f(s)=0$, and (ii) $f(x^4+y)=x^3f(x)+f(f(y))$ for all $x,y$ in ${\bf R}$. $\color{blue}\boxed{\textbf{Answer: }f\equiv x}$ $\color{blue}\boxed{\textbf{Proof:}}$ $\color{blue}\rule{24cm}{0.3pt}$ In $(ii) x\to 0:$ $$\Rightarrow f(y)=f(f(y))...(iii)$$Replacing $(iii)$ in $(ii):$ $$\Rightarrow f(x^4+y)=x^3f(x)+f(y)...(\alpha)$$In $(\alpha) x\to 1, y\to 0:$ $$\Rightarrow f(1)=f(0)+f(1)$$$$\Rightarrow f(0)=0$$$\color{red}\boxed{\text{If }\exists a\neq 0 / f(a)=0:}$ $\color{red}\rule{24cm}{0.3pt}$ In $(\alpha) x\to a:$ $$\Rightarrow f(y+a^4)=f(y)$$$$\Rightarrow f\text{ is periodic}$$By $(i) (\Rightarrow \Leftarrow)$ $\color{red}\rule{24cm}{0.3pt}$ $$\Rightarrow f(x)=0 \Leftrightarrow x=0...(\omega)$$In $(\alpha) y\to 0:$ $$\Rightarrow f(x^4)=x^3f(x)...(iv)$$Replacing $(iv)$ in $(\alpha):$ $$\Rightarrow f(x^4+y)=f(x^4)+f(y)...(\beta)$$In $(\beta) y\to f(y):$ $$\Rightarrow f(x^4+f(y))=f(x^4)+f(y)$$$$\Rightarrow f(x^4+f(y))=f(x^4+y)...(\theta)$$In $(\theta) y\to -x^4:$ $$\Rightarrow f(x^4+f(-x^4))=0$$By $(\omega):$ $$\Rightarrow f(-x^4)=-x^4$$$$\Rightarrow f(x)=x, \forall x\in\mathbb{R}^-_0...(1)$$In $(\beta) y\to -x^4-1:$ $$\Rightarrow -1=f(x^4)-x^4-1$$$$\Rightarrow f(x^4)=x^4$$$$\Rightarrow f(x)=x,\forall x\in\mathbb{R}^+...(2)$$By $(1)$ and $(2):$ $$\Rightarrow \boxed{f\equiv x}_\blacksquare$$$\color{blue}\rule{24cm}{0.3pt}$