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

Question

I read this question in the book 'Problem Solving Strategies ' by Arthur Engel. This question was asked in IMO of 1992. Here's how it goes

Find all the functions $ f:\mathbb{R} \mapsto \mathbb{R} $ that satisfy

$f[x^2+f(y)] = y +[f(x)]^2$ $ \ \ \ \ x,y \in \mathbb{R}$

The book simply states that it has no solution but it is not a sufficient answer. How can I prove that no such function exists ?

Answer 1

This partially answers the amended question with a different requirement: $$f(x^2+f(y)) = y+f(x)^2$$

Let $x=0$ to obtain $f(f(y)) = y + f(0)^2$. Let $y=0$ to obtain $f(x^2+f(0)) = f(x)^2$.

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Suppose $f$ has a fixed point: $y=f(y)$. Then $f(x^2+y) = y+f(x)^2$, so $f(y) = y+f(0)^2$ by letting $x=0$, and hence $f(0) = 0$. Therefore $f(x^2) = f(x)^2$ for all $x$, and $f(f(y)) = y$ for all $y$; in particular, $f$ is bijective.

But then $f(-x)^2 = f((-x)^2) = f(x^2) = f(x)^2$, so $f(-x) = \pm f(x)$; since $f$ is injective, that means $f(-x) = -f(x)$, so $f$ is odd.

Now letting $y=f(u)$, obtain $f(x^2+u) = f(u)+f(x)^2$, which is to say $f(x^2+u) = f(u) + f(x^2)$; therefore $f(z+u) = f(z) + f(u)$ whenever $z$ is nonnegative. By oddness of $f$, this also holds when both $z$ and $u$ are negative; so we have shown that $f$ is self-inverse, has $f(0) = 0$, has $f(x)^2 = f(x^2)$, and distributes over $+$. (These conditions imply the original requirement; but we are still operating under the assumption that $f$ has a fixed point.)

Since $f$ is odd, that means $f(x) = \sqrt{f(x^2)}$ when $x \geq 0$, and $-\sqrt{f(x^2)}$ when $x < 0$.

Noting then that $f(1)^2 = f(1)$, we have $f(1) = \pm 1$; but $f$ is nonnegative when given nonnegative input since $f(x)^2 = f(x^2)$, so $f$ is the identity on the rationals (by following the reasoning of my answer to the other version of this question). Therefore it is also the identity on the inductively defined set $X = \cup X_n$ where $X_1 = \mathbb{Q}, X_{n+1} = X_n \cup \{\sqrt{x}: x \in X_n\} \cup \{x+q : x \in X_n\} \cup \{x^2 : x \in X_n\}$. But (for example) $\pi$ is not in this set, and I think $f$'s value is not determined on $\pi$.

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If $f$ does not have a fixed point, I'm afraid I've got nowhere.

Answer 2

If "$[ \ ]$" is just a parenthesis, we can find all solutions! Start by taking $x=0$, so that we obtain $$ f(0^2+f(y)) = f(f(y)) = y+f(0) \quad \iff \quad f(0) = f(f(y))-y. $$ Then take $y=0$ to get $$ f(x^2+f(0)) = f(x^2). $$ Now let's plug $f(0)$ inside the latter expression $$ f(x^2) = f(x^2+f(0)) = f(x^2 -y +f(f(y)) = f(y) + f(x^2-y). $$ Note that the above expression holds only if $x^2-y\geq 0$, as it can be written as a square of a number.

Take $y=0$, so that we get $$f(x^2) = f(0)+f(x^2) \quad \Rightarrow \quad f(0)=0. $$ That is, $f(f(y))=y$. This can be translated as $f=f^{-1}$.

Answer 3

EDIT: this answer is answering the question originally stated ("$f(x^2+f(y)) = y+f(x^2)$"). The question has subsequently changed.

Partial answer here; in fact there are at least continuum-many solutions, though only two continuous ones.

Let $x=0$; then $f(f(y)) = y + f(0)$; in particular, $f(f(0)) = f(0)$. So:

$f$ does have a fixed point.

Let $y$ be a fixed point of $f$. Then $f(x^2+y) = y+f(x^2)$ for all $x$; in particular, $f(y) = y + f(0)$ by letting $x=0$; and so $y = y + f(0)$. So:

$f(0) = 0$.

Hence by letting $x = 0$, we have:

$$f(f(y)) = y$$ for all $y$.

In particular, $f$ is bijective.

Now fix $x$, and let $y = f(-x^2)$. We have $f(0) = f(x^2 + f(y)) = f(-x^2) + f(x^2)$; so $f(x^2) = -f(-x^2)$ for all $x$, and hence:

$f$ is odd.

Now, $f(x^2) + y = f(x^2 + f(y))$ so $f(x^2) + f(z) = f(x^2+z)$ by letting $y = f(z)$; in particular, if $x \geq 0$, then $f(x) + f(z) = f(x+z)$.

And of course, if both $x, z$ are less than $0$, we have $f(x) + f(z) = -(f(-x) + f(-z))$ by oddness, which is $-f(-x-z)$, which is $f(x+z)$ by oddness again. So:

$$f(x+z) = f(x) + f(z)$$ for all $x, z$; and $f$ is odd.

In fact, the conditions above ("$f$ is self-inverse, $f(0) = 0$, and $f$ distributes over $+$") are equivalent to the original condition.

Certainly if $f$ satisfies those conditions, then $g: x \mapsto -f(x)$ will also satisfy those conditions (and, indeed, $f: x \mapsto \pm x$ both work).

Another property: $f(nx) = n f(x)$ [an easy induction on $n$], and so $f\left(\frac{1}{n} x\right) = \frac{1}{n} f(x)$ and hence

$f\left(\frac{p}{q} x\right) = \frac{p}{q} f(x)$.

But consider the function $f: x \mapsto x$ if $x$ is not a rational multiple of $\sqrt{2}$ or $\sqrt{3}$; $q \sqrt{2} \mapsto q \sqrt{3}$, and $q \sqrt{3} \mapsto q \sqrt{2}$. This is certainly self-inverse, odd, and has $f(0) = 0$, but is not the identity. And indeed if $a, b$ are irrational numbers which are not rational multiples of each other, then $f: x \mapsto x$ if $x$ is not a rational multiple of either $a$ or $b$, and $a x \mapsto b x$, $b x \mapsto a x$ otherwise, is a function which satisfies the definition.

Answer 4

In the case that $ [ \cdot ] $ means the floor function (also denoted by $ \lfloor \cdot \rfloor $), you can show that there is no $ f $ satisfying $ f \big( \big\lfloor x ^ 2 + f ( y ) \big\rfloor \big) = y + \big\lfloor f \big( x ^ 2 \big) \big\rfloor $. Let $ x = 0 $ and you'll get $ f \big( \lfloor f ( y ) \rfloor \big) = y + \lfloor f ( 0 ) \rfloor $. This shows that $ f $ maps a subset of $ \mathbb Z $ onto $ \mathbb R $, which is imposible, since $ \mathbb Z $ is countable and $ \mathbb R $ is uncountable. The case is similar for the functional equation $ f \big( \big\lfloor x ^ 2 + f ( y ) \big\rfloor \big) = y + \lfloor f ( x ) \rfloor ^ 2 $, which corresponds with the IMO problem.