The function is not identically $0$. Let $x=a$ with $f(a)\ne0$ and $y=0$, so that $f(a)f(0)+f(a)=0$ implies $f(0)=-1$. Let $x=1$ and $y=-1$, so that $f(1)f(-1)-1=-1$ implies $f(1)f(-1)=0$. In the first case, consider $f(1)=0$. Setting $y=1$ and $x=z-1$ yields $f(z)=z-1$ for all $z$. In the second case, consider $f(-1)=0$. Setting $y=-1$ and $x=z+1$ yields $f(z)=-z-1$ for all $z$. Hence the only possibilties are $f(z)=z-1$ and $f(z)=-z-1$, and it is easily verified that they indeed work.

f(z)=z-1 and f(z)=-z-1

More than similar - identical (just take $g=-f$).

If we take $x$ = $-y$, then we get $$f(x)f(-x)=-x^2+1$$Is it right to assume from here that $$f(x)f(-x)=(1+x)(1-x)$$and hence $f(x) = 1 +x $ and $f(x) = 1- x$ or the other way around. Thanks

I don't think because f(0)=-1, you can prove that by taking (x,y)=(0,0)

JoelBinu wrote: If we take $x$ = $-y$, then we get $$f(x)f(-x)=-x^2+1$$Is it right to assume from here that $$f(x)f(-x)=(1+x)(1-x)$$and hence $f(x) = 1 +x $ and $f(x) = 1- x$ or the other way around. Thanks How do you conclude this? $f(x)f(-x)=(1+x)(1-x)$ and hence $f(x) = 1 +x $ and $f(x) = 1- x$

djb86 wrote: Find all functions $f:\mathbb{R}\to\mathbb{R}$ such that $f(x)f(y)+f(x+y)=xy$ for all real numbers $x$ and $y$. Let $P(x,y)$ the assertion of the given FE: $P(x,0)$ $$f(x)f(0)+f(x)=0 \implies f(x)=0 \; \text{or} \; f(0)=-1$$$P(-1,1)$ $$f(-1)f(1)-1=-1 \implies f(1)=0 \; \text{or} \; f(-1)=0$$If $f(1)=0$. $P(x-1,1)$ $$f(x-1)f(1)+f(x)=x-1 \implies f(x)=x-1$$If $f(-1)=0$. $P(x+1,-1)$ $$f(x+1)f(-1)+f(x)=-1-x \implies f(x)=-x-1$$Hence the solutions are: $\boxed{f(x)=x-1 \; \forall x \in \mathbb R}$ $\boxed{f(x)=-x-1 \; \forall x \in \mathbb R}$ Thus we are done

Since $f\not\equiv0$, $\exists j:f(j)\ne0$ $P(j,0)\Rightarrow f(0)=-1$ $P(1,-1)\Rightarrow f(-1)f(1)=0$ $f(1)=0\Rightarrow P(x-1,1)\Rightarrow\boxed{f(x)=x-1}$ $f(-1)=0\Rightarrow P(x+1,-1)\Rightarrow\boxed{f(x)=-x-1}$, which both work.

TuZo wrote: JoelBinu wrote: If we take $x$ = $-y$, then we get $$f(x)f(-x)=-x^2+1$$Is it right to assume from here that $$f(x)f(-x)=(1+x)(1-x)$$and hence $f(x) = 1 +x $ and $f(x) = 1- x$ or the other way around. Thanks How do you conclude this? $f(x)f(-x)=(1+x)(1-x)$ and hence $f(x) = 1 +x $ and $f(x) = 1- x$ Maybe he just guess more 2 sols

MathLuis wrote: djb86 wrote: Find all functions $f:\mathbb{R}\to\mathbb{R}$ such that $f(x)f(y)+f(x+y)=xy$ for all real numbers $x$ and $y$. Let $P(x,y)$ the assertion of the given FE: $P(x,0)$ $$f(x)f(0)+f(x)=0 \implies f(x)=0 \; \text{or} \; f(0)=-1$$$P(-1,1)$ $$f(-1)f(1)-1=-1 \implies f(1)=0 \; \text{or} \; f(-1)=0$$If $f(1)=0$. $P(x-1,1)$ $$f(x-1)f(1)+f(x)=x-1 \implies f(x)=x-1$$If $f(-1)=0$. $P(x+1,-1)$ $$f(x+1)f(-1)+f(x)=-1-x \implies f(x)=-x-1$$Hence the solutions are: $\boxed{f(x) \equiv 0 \; \forall x \in \mathbb R}$ $\boxed{f(x)=x-1 \; \forall x \in \mathbb R}$ $\boxed{f(x)=-x-1 \; \forall x \in \mathbb R}$ Thus we are done And the first sol doesn't fit the first FE.

IMOStarter wrote: MathLuis wrote: djb86 wrote: Find all functions $f:\mathbb{R}\to\mathbb{R}$ such that $f(x)f(y)+f(x+y)=xy$ for all real numbers $x$ and $y$. Let $P(x,y)$ the assertion of the given FE: $P(x,0)$ $$f(x)f(0)+f(x)=0 \implies f(x)=0 \; \text{or} \; f(0)=-1$$$P(-1,1)$ $$f(-1)f(1)-1=-1 \implies f(1)=0 \; \text{or} \; f(-1)=0$$If $f(1)=0$. $P(x-1,1)$ $$f(x-1)f(1)+f(x)=x-1 \implies f(x)=x-1$$If $f(-1)=0$. $P(x+1,-1)$ $$f(x+1)f(-1)+f(x)=-1-x \implies f(x)=-x-1$$Hence the solutions are: $\boxed{f(x) \equiv 0 \; \forall x \in \mathbb R}$ $\boxed{f(x)=x-1 \; \forall x \in \mathbb R}$ $\boxed{f(x)=-x-1 \; \forall x \in \mathbb R}$ Thus we are done And the first sol doesn't fit the first FE. Fixed, i was thinking in another FE while solving this lol. Sorry for the typo

$P(x,0) : f(x)f(0) + f(x) = 0 \implies f(0) = -1$. $P(-1,1) : f(-1)f(1) = 0$ so either $f(1) = 0$ or $f(-1) = 0$. Case $1 : f(1) = 0$ $P(x,1) : f(x+1) = x \implies f(x) = x-1$ Case $2 : f(-1) = 0$ $P(x,-1) = f(x-1) = -x \implies f(x) = -x - 1$ Now assume there exists $a,b$ such that $f(a) = a-1, f(b) = -b - 1$. $P(a,b) : -ab -a + b + 1 + f(a+b) = ab \implies f(a+b) = 2ab + a - b - 1$ which gives contradiction cause $f(a+b) = a + b - 1$ or $ -a - b -1$. Answers : $f(x) = x - 1, f(x) = -x - 1$

Claim: $f(0)=-1$. Proof: Suppose not. Then $P(0,0): f(0)^2+f(0)=0\implies f(0)\in \{0,-1\}\implies f(0)=0$. $P(x,0): f(x)=0$, but $f\equiv 0$ is not a solution $\blacksquare$ $P(1,-1): f(1)f(-1)-1=-1\implies f(1)f(-1)=0$. Case 1: $f(-1)=0$. $P(x,-1): f(x-1)=-x\implies \boxed{f(x)=-x-1},$ which works. Case 2: $(1)=0$. $P(x,1): f(x+1)=x\implies \boxed{f(x)=x-1}$, which works.