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Find all functions $f$ defined over real numbers to real numbers such that $f(f(y))+f(x-y)=f(xf(y)-x)$

My attempt: Set $x=y=0$ to get $f(f(0))=0$. It will be very helpful if I will able to find $f(0)$ but I failed to find it. I tried to check the injectivity of $f$ but wasn't able to check it.

Please help me.

Sufaid Saleel
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    Brief observation: Suppose $f(y) = 0$. Then by setting $x=0$, we get $f(0) + f(-y) = f(0)$. So $y$ is a root iff $-y$ is a root. Now instead suppose $f(x) = 0$. Then by setting $x=y$, we obtain $2f(0)=f(-x)$; but we already know $f(x) = 0 = f(-x)$, so $2f(0) = f(0)$ and hence (if there is any root at all) $f(0) = 0$. – Patrick Stevens Mar 22 '19 at 04:07
  • Okay, so $f(0)=0$.Now what to do next?We can also conclude that $f(x)=f(-x)$ – Sufaid Saleel Mar 22 '19 at 04:47
  • $f(0) = 0$ if there is a root. – Patrick Stevens Mar 22 '19 at 04:51

2 Answers2

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Let $P(x,y)$ be the assertion that $f(f(y))+f(x-y)=f(xf(y)-x)$. Now,

Observation 1: $P(0,0)\implies f(f(0))=0$.

Observation 2: Considering $P(0,f(0))$ and $P(f(0),f(0))$ we get $f(0)=0$.

Observation 3: Considering $P(x,0)$ we get $f$ is even function, i.e;$f(x)=f(-x)$ and finally $P(x,-x)$ and $P(x,x)$ implies $f(x)=0$.

So, $f(x)=0,\forall x\in\mathbb{R}$.

tarit goswami
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I tried to solve this problem with taking help from the above comments.Please check it. enter image description here

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I am tooo lazy to LaTeX it .Please don't mind!

Sufaid Saleel
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