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I was able to find that $f(0)=1$ by setting $y=0$ to have $f(x)f(0)=f(x)$. DIviding on both sides, I was left with $f(0)=1. I still don't know how I can find all the possible functions and how I can use this.

Cheez
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  • Seems like $f(x)=1+x$ and $f(x)=1-x$ are the solutions but I don't see how to show them yet. – N. S. May 27 '20 at 18:38
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    If you don't add some context (how did you find $f(0)=1$, say), your question will be closed, exactly as this one: https://math.stackexchange.com/questions/3687605/the-function-f-mathbbr-to-mathbbr-satisfies-fx-fy-fx-y-x#comment7578799_3687605 –  May 27 '20 at 18:43
  • What is the domain of $f?$ – Sahiba Arora May 27 '20 at 20:10
  • @SahibaArora The domain is all real numbers for both $x$ and $y$ – Cheez May 28 '20 at 16:44
  • @Cheez That information belongs in the question and not as a comment. – Sahiba Arora May 28 '20 at 17:00

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I'll assume you meant $f:\mathbb{R} \to \mathbb{R}$ $$P(x,y) \implies f(x)f(y)=f(x+y)+xy$$ $$P(x,0) \implies f(x)f(0)=f(x) \implies f(0)=1$$ Since $f(x)=0$ is not a solution, and we can replace $x$ by $a$ such that $f(a) \ne 0$, Now $$P(1,-1) \implies f(1)f(-1)=0$$ Case 1: $f(1)=0$ $$P(x,1) \implies f(x+1)+x=0 \iff f(x)=1-x$$ Case 2: $f(-1)=0$ $$P(x,-1) \implies f(x-1)-x=0 \iff f(x)=x+1$$

Anas A. Ibrahim
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