parmenides51 wrote: Find all functions $f : R \to R$ that satisfy $f(x-y)^2= f(x)^2 -2x f(y)+y^2$ for all $x,y \in R$ Let $P(x,y):f(x-y)^2= f(x)^2 -2x f(y)+y^2$ $P(x,0):f(x)^2=f(x)^2-2xf(0)+0\Leftrightarrow f(0)=0$ $P(x,x):0=f(x)^2-2xf(x)+x^2=(f(x)-x)^2\Leftrightarrow f(x)=x$ $(x-y)^2=x^2-2xy+y^2$ So $\boxed{f(x)=x}$ is the only solution.

The version on IMOMath states the left hand side as $f\left((x-y)^2\right)$ rather than $f(x-y)^2$. I solve that version. As above, let $P(x,y)$ denotes the given assertion. $P(0,0)$ yields $f(0)\in\{0,1\}$ and $P(x,x)$ yields $f(0)=(f(x)-x)^2$. In particular, for $f(0)=0$, we find $f(x)=x$ for all $x$. Assume now that $f(0)=1$. Inspecting $P(0,y)$ we find $f(y^2)=y^2+1$. In particular, $f(y)=y+1$ on $[0,\infty)$. We next extend this to entire real line. To that end, fix an $x>0$, and an arbitrary $y<0$ (and recall $f(x)=x+1$ as shown already). $f((x-y)^2)= (x-y)^2+1 = (x+1)^2 - 2xf(y)+y^2$. Clearing this up and then dividing by $2x$ yields immediately $f(y)=y+1$.

at first I have posted parmenides51 wrote: Find all functions $f : R \to R$ that satisfy $f(x-y)^2= f(x)^2 -2x f(y)+y^2$ for all $x,y \in R$ and had a typo, as grupyorum noticed grupyorum wrote: The version on IMOMath states the left hand side as $f\left((x-y)^2\right)$ rather than $f(x-y)^2$. therefore I have corrected it

The above solutions seem to have missed something... parmenides51 wrote: Find all functions $f : R \to R$ that satisfy $f((x-y)^2)= f(x)^2 -2x f(y)+y^2$ for all $x,y \in R$ $$f((x-y)^2)= f(x)^2 -2x f(y)+y^2...(\alpha)$$In $(\alpha) x=0:$ $$\Rightarrow f(y^2)=y^2+f(0)^2$$$$\Rightarrow f(x)=x+c,\text{ c is constant and }x \in \mathbb{R}^+_0...(\beta)$$In $(\alpha) x,y\ge 0:$ By $(\beta):$ $$\Rightarrow x^2-2xy+c=x^2+2xc+c^2-2xy-2xc$$$$\Rightarrow c^2=c$$$$\Rightarrow c\in (0,1)$$If $c=0:$ $$\Rightarrow f(x)=x, \forall x\in\mathbb{R}^+_0$$In $(\alpha):$ $$(x-y)^2\ge 0 \Rightarrow x^2-2xy=f(x)^2-2xf(y)...(\theta)$$In $(\theta) x>0:$ $$\Rightarrow x^2-2xy=x^2-2xf(y)$$$$\Rightarrow f(y)=y, \forall y\in\mathbb{R}$$If $c=1:$ $$\Rightarrow f(x)=x+1, \forall x\in\mathbb{R}^+_0$$In $(\alpha):$ $$(x-y)^2\ge 0 \Rightarrow x^2-2xy+1=f(x)^2-2xf(y)...(\omega)$$In $(\omega) x>0:$ $$\Rightarrow x^2-2xy+1=x^2+2x+1-2xf(y)$$$$\Rightarrow f(y)=y+1, \forall y\in\mathbb{R}$$ $$\Rightarrow f(x)=x \text{ and } f(x)=x+1 \text{ are the only solutions}_\blacksquare$$