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by Phlembac Adib Hasan » Tue Jul 02, 2013 7:53 pm
Here is further clarification of my previous solution as requested from a member.
Let $f(0)=c$. Note that $P(0,0)$ implies $f(c)=c^2$.
From $P(0,y)\Longrightarrow f(f(y))=(c+y^2)^2$. So $f$ is injective on $(c,\infty]$
$f(\sqrt c)=0$
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So $P(c,0)\Longrightarrow f(c^2+c)=c^4$
$P(\sqrt c, c)\Longrightarrow f(c+c^2)=(f(\sqrt c)+c^2)^2$
Since $c$ is non-negative, So from the last two lines, $f(\sqrt c)=0$.
Building infinite sequence:
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Define $c_0=\sqrt c$ and $c_{i+1}=c_{i}^2+c$ for $i>0$. If $c>0$, this sequence is obviously increasing and unbounded. From $P(\sqrt c,0)$, $f(c_1)=0$.
And notice that if $f(c_i)=0$ then $P(c_i,0)\Longrightarrow f(c_{i+1})=f(c_i^2+c)=f(c_i)^2=0$. But $f$ is injective on $(c,\infty]$ which leads to a contradiction. So $c=0$.
Therefore from $P(x,0)$, $f(x^2)=f(x)^2$.
So $f(x^2+y^2)=(f(x)+f(y))^2=f(x)^2+2f(x)f(y)+f(y)^2=f(x^2)+w\ge f(x^2)$. So for all $x,y\ge 0$, $f(x+y)\ge f(x)$ i.e. $f$ is non-decreasing on non-negative reals.