Here it is.

Nice problem! Amir Hossein wrote: Let $n\geq 2$ be a given integer. Find all functions $f:\mathbb{R}\rightarrow \mathbb{R}$ such that \[f(x-f(y))=f(x+y^n)+f(f(y)+y^n), \]for all $x,y \in \mathbb{R}$. Answer. $f \equiv 0$ and $f(x)=-x^n$ for all $x \in \mathbb{R}$ are the only solutions. It is easy to verify they work. Now we prove that any good $f$ must be one of them. Suppose $f$ is not the zero function. Let $f(0)=2C$. Plug $x=f(y)$ so $f(f(y)+y^n)=C$ for all $y \in \mathbb{R}$. Claim. Suppose $f \not \equiv 0$, then $f$ is not a periodic function. (Proof) Suppose $f$ has a period $\ell$. Plug $y \mapsto (y+\ell)$ so $$f(x+y^n)=f(x+(y+\ell)^n)$$for all $x, y \in \mathbb{R}$. So $(y+\ell)^n-y^n$ is a period. It follows that $f$ has a period $N$ for all sufficiently large numbers $N$. Thus $f$ is the constant function, so $f \equiv 0$, a contradiction! $\blacksquare$ Now we claim that $f$ is either injective or an even function. Suppose we can find $a \ne b$ with $f(a)=f(b)$. If $a^2 \ne b^2$, then plug $y=a$ and $y=b$ to obtain $a^n-b^n$ as a period of $f$. However, the previous lemma then shows $f \equiv 0$. So $a^2=b^2$ hence $n$ even. But then $$f(x-f(y))=f(x+y^n)+C=f(x-f(-y))$$for $f(y)-f(-y)$ is a period. Again our lemma then proves $f$ is even. Now if $f$ is even, we simply put $x \mapsto -x$ so $$f(x+f(y))=f(x-y^n)+C$$for all $x,y \in \mathbb{R}$. Plug $x=y^n$ so $f(0)=0$ hence $C=0$. Now $y^n+f(y)$ is a period; so the lemma just shows $f(y)=f(0)-y^n=-y^n$ for all $y \in \mathbb{R}$. Finally, if $f$ is injective, then $f(f(y)+y^n)=f(f(0))=C$ so $f(y)=f(0)-y^n$ for all $y \in \mathbb{R}$. Plugging it back, we see $f(0)=0$ and $f(x)=-x^n$ for all $x \in \mathbb{R}$ as the other solution.

anantmudgal09 wrote: Nice problem! Amir Hossein wrote: Let $n\geq 2$ be a given integer. Find all functions $f:\mathbb{R}\rightarrow \mathbb{R}$ such that \[f(x-f(y))=f(x+y^n)+f(f(y)+y^n), \]for all $x,y \in \mathbb{R}$. Answer. $f \equiv 0$ and $f(x)=-x^n$ for all $x \in \mathbb{R}$ are the only solutions. It is easy to verify they work. Now we prove that any good $f$ must be one of them. Suppose $f$ is not the zero function. Let $f(0)=2C$. Plug $x=f(y)$ so $f(f(y)+y^n)=C$ for all $y \in \mathbb{R}$. Claim. Suppose $f \not \equiv 0$, then $f$ is not a periodic function. (Proof) Suppose $f$ has a period $\ell$. Plug $y \mapsto (y+\ell)$ so $$f(x+y^n)=f(x+(y+\ell)^n)$$for all $x, y \in \mathbb{R}$. So $(y+\ell)^n-y^n$ is a period. It follows that $f$ has a period $N$ for all sufficiently large numbers $N$. Thus $f$ is the constant function, so $f \equiv 0$, a contradiction! $\blacksquare$ Now we claim that $f$ is either injective or an even function. Suppose we can find $a \ne b$ with $f(a)=f(b)$. If $a^2 \ne b^2$, then plug $y=a$ and $y=b$ to obtain $a^n-b^n$ as a period of $f$. However, the previous lemma then shows $f \equiv 0$. So $a^2=b^2$ hence $n$ even. But then $$f(x-f(y))=f(x+y^n)+C=f(x-f(-y))$$for $f(y)-f(-y)$ is a period. Again our lemma then proves $f$ is even. Now if $f$ is even, we simply put $x \mapsto -x$ so $$f(x+f(y))=f(x-y^n)+C$$for all $x,y \in \mathbb{R}$. Plug $x=y^n$ so $f(0)=0$ hence $C=0$. Now $y^n+f(y)$ is a period; so the lemma just shows $f(y)=f(0)-y^n=-y^n$ for all $y \in \mathbb{R}$. Finally, if $f$ is injective, then $f(f(y)+y^n)=f(f(0))=C$ so $f(y)=f(0)-y^n$ for all $y \in \mathbb{R}$. Plugging it back, we see $f(0)=0$ and $f(x)=-x^n$ for all $x \in \mathbb{R}$ as the other solution.

anantmudgal09 wrote: Nice problem! Amir Hossein wrote: Let $n\geq 2$ be a given integer. Find all functions $f:\mathbb{R}\rightarrow \mathbb{R}$ such that \[f(x-f(y))=f(x+y^n)+f(f(y)+y^n), \]for all $x,y \in \mathbb{R}$. Answer. $f \equiv 0$ and $f(x)=-x^n$ for all $x \in \mathbb{R}$ are the only solutions. It is easy to verify they work. Now we prove that any good $f$ must be one of them. Suppose $f$ is not the zero function. Let $f(0)=2C$. Plug $x=f(y)$ so $f(f(y)+y^n)=C$ for all $y \in \mathbb{R}$. Claim. Suppose $f \not \equiv 0$, then $f$ is not a periodic function. (Proof) Suppose $f$ has a period $\ell$. Plug $y \mapsto (y+\ell)$ so $$f(x+y^n)=f(x+(y+\ell)^n)$$for all $x, y \in \mathbb{R}$. So $(y+\ell)^n-y^n$ is a period. It follows that $f$ has a period $N$ for all sufficiently large numbers $N$. Thus $f$ is the constant function, so $f \equiv 0$, a contradiction! $\blacksquare$ Now we claim that $f$ is either injective or an even function. Suppose we can find $a \ne b$ with $f(a)=f(b)$. If $a^2 \ne b^2$, then plug $y=a$ and $y=b$ to obtain $a^n-b^n$ as a period of $f$. However, the previous lemma then shows $f \equiv 0$. So $a^2=b^2$ hence $n$ even. But then $$f(x-f(y))=f(x+y^n)+C=f(x-f(-y))$$for $f(y)-f(-y)$ is a period. Again our lemma then proves $f$ is even. Now if $f$ is even, we simply put $x \mapsto -x$ so $$f(x+f(y))=f(x-y^n)+C$$for all $x,y \in \mathbb{R}$. Plug $x=y^n$ so $f(0)=0$ hence $C=0$. Now $y^n+f(y)$ is a period; so the lemma just shows $f(y)=f(0)-y^n=-y^n$ for all $y \in \mathbb{R}$. Finally, if $f$ is injective, then $f(f(y)+y^n)=f(f(0))=C$ so $f(y)=f(0)-y^n$ for all $y \in \mathbb{R}$. Plugging it back, we see $f(0)=0$ and $f(x)=-x^n$ for all $x \in \mathbb{R}$ as the other solution. Please can you explain how $$a^n-b^n$$is a period?

Let $P(x,y)$ denote the assertion $f(x-f(y))=f(x+y^k)+f(f(y)+y^k).$ We claim the only solutions which work are $f(x)\equiv -x^k$ and $f\equiv 0.$ $P(f(x),x)$ implies $f(x^k+f(x))=f(0)/2.$ Then combining $P(x-y^k-f(y),z^k+f(z))$ and $P(x-z^k-f(z),z^k+f(z))$ implies $f(x+a)=f(x)$ where $a=y^k+f(y)-z^k-f(z).$ If $a=0$ then we get our first solution. Consider $a\neq 0.$ Then taking $y=z+a$ implies $f$ is constant, our second solution.

@above how does $P(f(x),x)$ imply $f(x^k + f(x)) = f(0)/2$?. I think you made a mistake because $P(f(x),x)$ gives $f(f(x) - f(x)) = f(f(x) + x^k) + f(f(f(x)) + x^k)$ which simplifies to $f(0) = f(f(x) + x^k) + f(f(f(x)) + x^k)$ and you forgot an $f(x)$ i think.

straight wrote: @above how does $P(f(x),x)$ imply $f(x^k + f(x)) = f(0)/2$?. I think you made a mistake because $P(f(x),x)$ gives $f(f(x) - f(x)) = f(f(x) + x^k) + f(f(f(x)) + x^k)$ which simplifies to $f(0) = f(f(x) + x^k) + f(f(f(x)) + x^k)$ and you forgot an $f(x)$ i think. No $P(f(x),x)\implies f(f(x)-f(x))=f(f(x)+x^k)+f(f(x)+x^k)$ $\implies f(0)=2f(f(x)+x^k)\implies f(f(x)+x^k)=f(0)/2$

Solution: The answer is $f\equiv 0$ or $f(X)=-X^n$. Set $K:=\{f(y)+y^n\mid y\in \mathbb{R}\}$. From the condition, we know that $$f(x)-f(x+s)=f(s) \qquad \forall (x,s)\in \mathbb{R}\times K$$Take $x=0$, we have $f(s)=f(0)/2$ for any $s\in K$. Now we have the following two cases. Case 1: $K=\{0\}$ Obviously $f(x)=-x^n$. Case 2: There is a nonzero element in $K$. In this case, suppose $0\neq h\in K$. Then $$(f(y+h)+(y+h)^n)-(f(y)+y^n)=-f(h)+(y+h)^n-y^n$$is a polynomial of degree $n-1\geq 1$ in variable $y$. This indicates that there exists a constant $M>0$, such that for any real number $T\geq M$, there are two elements $s_1,s_2$ in $K$, $s_1-s_2=T$. Thus we have $$f(x)=f(x-s_1)+f(s_1)=f(x-s_1+s_2)+f(s_1)-f(s_2)=f(x-s_1+s_2)$$The above shows that $f(x)=f(x+T)$ for any $T\geq M$, from here we conclude that f is constant, thus $f\equiv 0$. Done!