Indian national mathematical olympiad

Find all functions f: $\mathbb { R }\implies\mathbb{ R }$ such that $f(x^2+yf(z)) = x f(x)+z f(y)$ , for all x , y , z in $\mathbb { R }$

Solution :Let $P(x,y,z)$ be the assertion $f(x^2+yf(z))=xf(x)+zf(y)$

$P(0,0,0)$ $\implies$ $f(0)=0$

$P(0,1,xf(x))$ $\implies$ $f(f(xf(x)))=xf(x)f(1)$
$P(0,x,x)$ $\implies$ $f(xf(x))=xf(x)$ and so $f(f(xf(x)))=xf(x)$ and so $xf(x)(f(1)-1)=0$

So, either $f(x)=0$ $\forall x\ne 0$, which gives the solution $\boxed{\text{S1 : }f(x)=0\text{  }\forall x}$
Either $f(1)=1$ and so $P(0,1,x)$ $\implies$ $f(f(x))=x$ and $f(x)$ is bijective.

Then $P(0,x,f(y))$ $\implies$ $f(xy)=f(x)f(y)$ and $f(x)$ is multiplicative.
So $P(x,0)$ $\implies$ $f(x)^2=f(x^2)=xf(x)$ and so $f(x)=x$ $\forall x\ne 0$, still true when $x=0$

And so $\boxed{\text{S2 : }f(x)=x\text{  }\forall x}$ which indeed is a solution

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