IMO-1992-2

[Phlembac Adib Hasan]

Find all functions $f:\mathbb{R}\rightarrow \mathbb{R}$ such that \[f(x^2+f(y))=y+[f(x)]^2\]

[shehab ahmed]

Here are the steps of my solution.
1.Put $x=0$ to deduce that $f$ is surjective.
2.Let,$f(a)=f(b)$.Now add $a^2$ in both sides and use the given equation to conclude that f is injective.
3.Let,$f(0)=c$,puting $x=0$ we have $f(f(y))=y + c^2$
4.as $f$ is surjective,there is a $k$ such that f(k)=0.Put $x=k$ and y=$f(y)$.Then use equation in the 3rd step and injectivity of $f$ to deduce that $c=0$
So the equation in the 3rd step reduces to $f(f(y))=y$
5.Now put $x=y$ and u will get an equation.In that equation put $x=f(x)$ to conclude that $f(x)^2=x^2$
6.let there are $c,d$ such that $f(c)=c$ and $f(d)=-d$,now put $x=c$ and $y=d$ and use the fact that $f(0)=0$ to deduce that for all real $x$ either $f(x)=x$ or $f(x)=-x$.
Now,checking kills the 2nd case

[shehab ahmed]

দুঃখিত,আমার চতুর্থ স্টেপের পর ভুল আছে।

[shehab ahmed]

After 4th step,
5.put $y=0$ to get $f(x^2)=f(x)^2$
6.Now put $y=f(x)$ and use the equation in the 5th step to get a Cauchy's equation.
Hence,$f(x)=cx$ for some constant $c$.Checking ensures that $c=1$

[*Mahi*]

6.Now put $y=f(x)$ and use the equation in the 5th step to get a Cauchy's equation.
Hence,$f(x)=cx$ for some constant $c$.Checking ensures that $c=1$

$f$ is $\mathbb R \mapsto \mathbb R$. So you can not use cauchy equation without proving some additional properties of $f$

[shehab ahmed]

হায়রে,মোবাইল থেকে কিচ্ছু বুঝতে পারছি না।আন্দাজের উপর ডোমেন এবং কো ডোমেন ধরে নিয়ে অঙ্ক করা শুরু করেছি।একটু সোজা বাংলায় ডোমেন এবং কোডোমেন কী তা বলে দিলে ভাল হত

[shehab ahmed]

ও আচ্ছা।ধন্যবাদ মাহি।এটা আমার মনে ছিল না।
additional property of f:
$f(x)^2=f(x^2)$ now if $x$ is non negative, then putting $x=x^{\frac {1}{2}}$ we have $f(x)$ is non negative.So we can use cauchy.

[*Mahi*]

additional property of $f$:
$f(x)^2=f(x^2)$ now if $x$ is non negative then putting $x=x^{1/2}$ we have $f(x)$ is non negative.So we can use cauchy

That should be enough. But please never forget it in contests (which may cost you severely ).

[Nadim Ul Abrar]

$p(x,y) : f(x^2+f(y))=y+(f(x))^2$

Now $p(0,y) : f(f(y))=y+(f(0))^2$
that imply $f$ is bijective .

Now consider $p(-x,y)$ .
$f(x)= \pm f(-x)$ .

Using injectivity we can say that , $f(x) \neq f(-x)$ for all $x$ .
So $f(x)=-f(-x)$ ...(0).

using surjectivity ,
we can find a '$c$' such that $f(c)=0$ .

$f(-c)=-f(c)=0=f(c)$ .
So injectivity say's that $c=-c$ ,$ c=0$

Now p(x,0): $f(x^2)=f(x)^2$
put $x=1$ . $f(1)=(f(1))^2$
or $f(1)=1$

So $p(x,1): f(x^2+1)=1+(f(x))^2=1+f(x^2)$ ***

If $z$ be a positive integer then ,
for all positive integer $k \leq z^2 , z^2-k$ is square of a real number

Now $p(z,1) : f(z^2+1)=1+f(z^2)=1+f(z^2-1+1)=1+f(z_1^2+1)=2+f(z_1^2)=...$ ($z^2-1=z_1^2$ , $z_1$ is real )
proceeding through this way we will find
$f(z^2+1)=f(z^2-z^2)+z^2+1=z^2+1$ .

so $f(z^2)=z^2$ .

Now consider $p(z,y): f(z^2+f(y))=y+z^2$ ...(1).
set $f(y)+z^2-1=a_1^2$
$f(a_1^2+1)=1+f(a_1)^2$
....
Thus get $f(z^2+f(y))=f(y)+z^2$ ...(2)

using (0),(1),(2) f(x)=x

[Phlembac Adib Hasan]

Here is my proof:

Let $p(x,y\Rightarrow f(x^2+f(y))=y+[f(x)]^2$.
Now $p(0,y)$ implies $f(x)$ is bijective and $f(f(x))=x+[f(0)]^2............(i)$.
From injectivity and $p(-x,y)$ we can say $f(x)=-f(-x).....(ii)$
From surjectivity let $f(c)=0$.Now $p(c,c)$ gives $f(c^2)=c\Rightarrow f(f(c^2))=f(c)=0\Rightarrow c^2+f(0)^2=0(\text {From}\; \; (i)$.So $c=f(0)=0$.
So $(i)$ becomes $f(f(x))=x....(iv)$
Now $p(x,0)$ gives $f(x^2)=[f(x)]^2\ge 0....(v)$
Therefore the main equation becomes $f(x^2+f(y))=f(x^2)+f(f(y))$.Let $m=x^2$ and $n=f(y)$.So it becomes $f(m+n)=f(m)+f(n)\forall m\in \mathbb {R}_+\cup 0,n\in \mathbb {R}.....(vi)$
Now we'll prove that $f(x+y)=f(x)+f(y)\forall x,y\in \mathbb {R}$
If $x,y\ge 0$ or $x\ge 0,y<0$ or $x<0,y\ge 0$, then it's trivial from $(vi)$.So one case is remaining to show when $x,y<0$.
Let $x=-p$ and $y=-q$ with $p,q>0$.So $f(x+y)=f(-(p+q))=-f(p+q)=-f(p)-f(q)=f(-p)+f(-q)=f(x)+f(y)$.
Also remember $f(x)\ge 0\forall x\in \mathbb {R}_+\cup 0$.So $f(x)=\eta x$.By putting this value in the main equation we find that $\eta =1$.So $f(x)=x\forall x\in \mathbb {R}$.It's also easy to check that this is a right solution.