## F.E. (2012 Croatian TST)

For discussing Olympiad Level Algebra (and Inequality) problems
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### F.E. (2012 Croatian TST)

Find all $f:R\rightarrow R$ satisfying
$f(x^2+f(y))=(f(x)+y^2)^2$

My Solution Sketch
i)$f(x)=f(-x)$
ii)$f(x)\geq 0$
iii) $f(x)=f(y)$ implies $x=\pm y$
iv)$P(x,\sqrt(f(y)))$ and swap $(x,y)$
v) $f(x^2+y^2)=(f(x)+f(y))^2$
vi)$f(0)=0$ or $f(0)=.5$ (Doesn't satisfy the equation)
vii)$f(x^2)=(f(x))^2$
viii) $\sqrt(f(x))=g(x)$
ix) $g(x+y)=g(x)+g(y)$ and solve it.
You spin my head right round right round,
When you go down, when you go down down......
(-$from$ "$THE$ $UGLY$ $TRUTH$" )

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### Re: F.E. (2012 Croatian TST)

Till (v) my solution is same. I proved $c=f(0)=0$ in some other way: I showed $f(\sqrt {c})=0$ and then built an infinite increasing sequence $c_i$ of positive reals such that $f(c_i)=0$. Since $f$ is injective on $(c,\infty ]$, it obviously leads to a contradiction. And then $f(x^2)=f(x)^2$.
vi) Now from (v), $f$ is non-decreasing.

vii) $\boxed{\text{Claim:}}$ If $f(a)=b^2$ and $f(b)=a^2$, then $a=b$.

$\boxed{\text{Proof:}}$ WLOG we may assume $a\ge b\Longrightarrow f(a)\ge f(b)\Longrightarrow b^2\ge a^2\Longrightarrow a=b$.
viii) $P(x,x)\Longrightarrow f(x^2+f(x))=(x^2+f(x))^2$ i.e. $\exists z:f(z)=z^2$.

ix) $P(x,z)\Longrightarrow f(x^2+z^2)=(z^2+f(x))^2$
x) $P(z,x)\Longrightarrow f(z^2+f(x))=(z^2+x^2)^2$

Take $a=x^2+z^2$ and $b=z^2+f(x)$. Now from claim (vii) $a=b\Longrightarrow x^2+z^2=z^2+f(x)\Longrightarrow f(x)=x^2\forall x\in \mathbb R$.

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### Re: F.E. (2012 Croatian TST)

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 $(c,\infty]$

$f(\sqrt c)=0$
============
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:
======================
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.