Let $\mathbb{R}$ be the set of real numbers. Determine all functions $f: \mathbb{R}\longrightarrow \mathbb{R}$ so that the equality $$f(x+yf(x+y)) +xf(x)= f(xf(x+y+1))+y^2$$is true for any real numbers $x,y$.
Problem
Source: Pan-American Girls’ Mathematical Olympiad, P3
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jasperE3
06.10.2021 00:39
Let $P(x,y)$ denote the assertion. $P(-1,0)\Rightarrow f(-f(0))=0$ $P(0,-f(0))\Rightarrow f(0)=0$ $P(0,1)\Rightarrow f(f(1))=1$ $P(0,f(1))\Rightarrow f(1)^2=1$ Case 1: $f(1)=1$ $P(x,-x)\Rightarrow xf(x)=x^2\Rightarrow\boxed{f(x)=x}$ since $f(0)=0$. This is a solution. Case 2: $f(1)=-1$ Then $f(-1)=1$. $P(x,-x-1)\Rightarrow xf(x)=x^2+2x\Rightarrow f(1)=3$, contradiction.
daniel73b
07.10.2021 12:53
I have a slightly different solution. Setting $x=-1$, $y=0$ we have $f(-f(0))=0$, or real $z$ exists such that $f(z)=0$. Take $x=0$ and $y=z$, or $f(0)=f(0)+z^2$, hence $f(x)=0$ iff $x=0$. Setting $x=0$, $y=1$ we have $f(f(1))=1$, or real $u$ exists such that $f(u)=1$. Setting $x=u-1$, $y=0$ yields $f(u-1)+(u-1)f(u-1)=f((u-1)f(u))=f(u-1)$, or $(u-1)f(u-1)=0$. If $u\neq1$ then $f(u-1)=0$ for $u-1=0$, contradiction. Hence $u=1$, and $f(x)=1$ iff $x=1$. Setting finally $y=-x$ yields $f(x)+xf(x)=f(x)+x^2$, or $xf(x)=x^2$ for all real $x$. Thus $f(x)=x$ for all nonzero $x$ and $f(0)=0$, and the only possible solution is $f(x)=x$. Substitution into the condition yields both sides equal to $x^2+y^2+xy+x$, and we are done.
BarisKoyuncu
12.10.2021 20:05
Let $P(x,y)$ be the assertion. $P(-1,0)$ gives $f(-f(0))=0$. $P(0,y)$ gives $f(yf(y))=f(0)+y^2$. Take $y=-f(0)$ in this equation. $f(0)=f(0)+f(0)^2\Rightarrow f(0)=0$. Then, $f(xf(x))=x^2\Rightarrow f(f(1))=1$. $P(1,-1)$ gives $f(1)+f(1)=f(f(1))+1=2\Rightarrow f(1)=1$. $P(x,-x)$ gives $f(x)+xf(x)=f(x)+x^2\Rightarrow xf(x)=x^2\Rightarrow f(x)=x$.
Humberto_Filho
12.10.2021 22:56
Let P(x,y) denote the assertion. Start with the basic and usual P(0,x) and P(x,0) Leaving us with $f(xf(x)) = f(0) + x^2 $ FACT 1 and $f(x) + xf(x) = f(xf(x+1))$ FACT 2 using x=-1, on FACT 2, we get $f(-1) - f(-1) = f(-f(0))$ so, $f(-f(0)) = 0$ Letting y = -f(0) and FACT 1, we get $f(-f(0) f(-f(0)) = f(0) + f(0)^2$ so, f(0) = 0 . And leting x=1 on FACT 1, we get that $f(f(1)) = 1 $ Letting x=1, y=-1, we get $f(1) + f(1) = f(f(1)) + 1 $ so $f(1)=1$ FACT 3 and $f(0) = 0$ FACT 4 Using FACT 3 and 4 on the original equation with x=x and y=-x, we get $f(x-x.f(0) + xf(x) = f(x.f(1)) + x^2 $ so $xf(x) = x^2 $ and f(x) = x cuz x isn't 0. The unique solution is f(x)=x.