I read this question in the book 'Problem Solving Strategies ' by Arthur Engel. This question was asked in IMO of 1992. Hers'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 ?

Snapshot of the question

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    $\begingroup$ Is $[?]$ meant to denote the nearest integer? $\endgroup$ – Pedro Tamaroff Jan 31 '18 at 12:15
  • $\begingroup$ Is $f$ differentiable? $\endgroup$ – Mostafa Ayaz Jan 31 '18 at 12:27
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    $\begingroup$ If square brackets are simply brackets, then $f$ may be the identity, of course, and I have a collection of strong properties of $f$ in that case. $\endgroup$ – Patrick Stevens Jan 31 '18 at 12:28
  • $\begingroup$ I think it’s simple square brackets but I’ve still posted a picture of the question $\endgroup$ – Samyak Jha Jan 31 '18 at 14:07
  • $\begingroup$ The IMO question was $\dots y+f(x)^2$ not $\dots y+f(x^2)$ and it does have a solution. $\endgroup$ – almagest Jan 31 '18 at 16:10

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 $f(x^2+f(0)) = f(x)^2$.

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 $\pi$.

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

  • $\begingroup$ In the case that there is a fixed point, the function is increasing, since it takes positive values at positive points. Together with the fact that it's identity on rationals, it shows that the function must be equal to identity everywhere. $\endgroup$ – Mohsen Shahriari Feb 2 '18 at 0:50

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 $f=f^{-1}$.

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    $\begingroup$ Could you explain more why $f(x^2-y+f(f(y)))=f(y)+x^2-y$? Thanks in advance. $\endgroup$ – awllower Jan 31 '18 at 12:34
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    $\begingroup$ I think the reasoning becomes neater if you use the following two lines at the front: "Let $x=0$; then $f(f(y)) = y + f(0)$; in particular, $f(f(0)) = f(0)$. So $f$ does have a fixed point. Now 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 $f(0) = 0$." Then the numbers become neater throughout, and in particular the step $f(y) = y+c$ to $f(y) = y$ is trivial, since $y = f(f(y)) = y+2c$. $\endgroup$ – Patrick Stevens Jan 31 '18 at 12:37
  • $\begingroup$ I just correct it, sorry. $\endgroup$ – Tommaso Seneci Jan 31 '18 at 12:51
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    $\begingroup$ I now don't like your final statement. You say "the only self-inverse maps are $x$ and $-x$", but that's not true. $\endgroup$ – Patrick Stevens Jan 31 '18 at 13:10
  • $\begingroup$ If it is smooth than we can prove that, using the derivative of the inverse function theorem, that these are only the functions $\pm x +b$, $b\in\mathbb{R}$, if it not smooth I don't know. I'll think about it. $\endgroup$ – Tommaso Seneci Jan 31 '18 at 13:18

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.


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.

  • $\begingroup$ But it’s given that the function maps from Real to Real set $\endgroup$ – Samyak Jha Feb 1 '18 at 2:12
  • $\begingroup$ Note that the left side of the equation is the value of $ f $ at an integer. Changing $ y $ through all over the real line, the left side takes countably many values, while the right side takes uncountably many. $\endgroup$ – Mohsen Shahriari Feb 2 '18 at 0:42

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