Problem about functional equations: $ f \left( x ^ 2 + y \right) = f ( x ) ^ 2 + \frac { f ( x y ) } { f ( x ) } $

Question

This problem was taken from the Bulgarian selection team test for the 47th IMO and appeared in a Chinese magazine, I came across it in my own training.

http://www.math.ust.hk/excalibur/v10_n4.pdf

The problem is as follows:

Consider $ f : \mathbb R ^ * \to \mathbb R ^ * $ where $ \mathbb R ^ * $ is the set of non-zero real numbers. Find all such functions that satisfy the following functional equation for all $ x $ and $ y $ in $ \mathbb R ^ * $ with $ y \ne - x ^ 2$: $$ f \left( x ^ 2 + y \right) = f ( x ) ^ 2 + \frac { f ( x y ) } { f ( x ) } $$

I suspect the only function is the identity, but the problem has proved itself to be too difficult for me, I've pondered about it for various hours. Any hint or solution is welcome! Thanks.

Answer 1

This might not be the optimal approach... but it attempts to avoid reasoning based on real analysis (well, apart from the very definition of irrational numbers).

Determine the value of $f(1)$

Let $f(1)=A$. Then, for any $y$ different from $0$ and $(-1)$ we have $$f(1+y)=f(1)^2 + f(y)/f(1) = A^2 + f(y)/A$$ which allows us to express the following quantities: $$\begin{eqnarray} f(2) & = & A^2 + 1 \\ f(3) & = & A^2 + A + \frac{1}{A} \\ f(4) & = & A^2 + A + 1 + \frac{1}{A^2} \\ f(5) & = & A^2 + A + 1 + \frac{1}{A} + \frac{1}{A^3} \\ \end{eqnarray}$$

At the same time, we have $f(x^2+1)=f(x)^2+f(x)/f(x)=f(x)^2+1$. This gives us $$f(5)=f(2)^2+1$$ Equating the two expressions for $f(5)$ and substituting for $f(2)$ gives us $$(A^2+1)^2 + 1 = A^2+A+1+\frac{1}{A} + \frac{1}{A^3}$$ which simplifies to $$\begin{eqnarray}0 & = & A^7 + A^5 - A^4 + A^3 - A^2 - 1 \\ & = & (A-1)(A^2-A+1)(A^2+A+1)^2\end{eqnarray}$$ whose only real root is $A=1$. Thus, $f(1)=1$ and we have $f(1+y)=1+f(y)$ for all $y$ other than $0$ and $(-1)$. In particular, this means that $f(k)=k$ for any positive integer $k$.

Show additive property for numbers of same sign

Now, consider real numbers $u,v$ of the same sign and these two equalities: $$\begin{eqnarray} f(v^2 + u/v) & = & f(v)^2 + f(u)/f(v) \\ f(v^2 + (u/v + 1)) & = & f(v)^2 + f(u+v)/f(v) \\ \end{eqnarray}$$ We can apply the equality $f(1+y)=1+f(y)$ since both $v^2$ and $u/v$ are positive to obtain $$f(u+v)=f(u)+f(v)$$

Note: This equation is very similar to Cauchy's functional equation; a quite interesting problem of its own. The equality given in our problem is stronger than "just" plain additivity, though.

Determine values of $f$ on positive rationals

It's quite easy to see that this equality implies that $f\left(\frac{p}{q}\right)=\frac{p}{q}$ for any positive integers $p,q$; one simply expands $$p=f(p)=f\left(\frac{p}{q}\right)+f\left(\frac{p}{q}\right)+\ldots+f\left(\frac{p}{q}\right)=q\cdot f\left(\frac{p}{q}\right)$$

Show that $f$ is increasing on positive reals

For $y>0$, we can write $$\begin{eqnarray} f(y) = f(y)+1-1 & = & f(y+1)-1 \\ & = & f\left((\sqrt{y})^2+1\right)-1 = \\ & = & f\left(\sqrt{y}\right)^2+1-1 \\ & = & f\left(\sqrt{y}\right)^2>0 \end{eqnarray}$$

Thus, for any positive reals $u>v>0$, we get $$f(u)=f(v+(u-v))=f(v)+f(u-v)>f(v)$$

Determine the values of $f$ for positive reals

Let's show that $f(x)=x$ for $x>0$. If that wasn't the case, we'd have either $f(x)>x$ or $0<f(x)<x$. In the first case, we could find rational number $z$ with $f(x)>z>x$. First inequality implies $f(x)>f(z)$ since $f(z)=z$, while the second one implies $f(z)>f(x)$, a contradiction. The other case can be treated in the same way. Thus, $f(x)=x$.

Alternate approach is to realize that the Dedekind cut corresponding to irrational $x>0$ must be the same as the Dedekind cut corresponding to $f(x)$.

Finish the proof by considering the negative reals

If $x<0$ is non-integer, we can simply use the equality $f(1+y)=1+f(y)$ to show that $f(x)=x$; just by adding $1$ until we reach a positive number. For negative integers, we can bootstrap the process by considering $f(-1)=f\left(-\frac{1}{2}\right)+f\left(-\frac{1}{2}\right)=(-1)$ and do the same by reaching $(-1)$.

The only function satisfying the given condition is $f(x)=x$ Q.E.D.

Answer 2

Here is a solution to the actual problem (with a cheap ending).

First because $f(x^2 + y) = f((-x)^2 + y)$, taking $y = 1$ we have $$ f(x)^2 + 1 = f(-x)^2 + 1 $$ so $f(x)^2 = f(-x)^2$, i.e. $|f(x)| = |f(-x)|$. Thus going back to general $y$, $$ f(x)^2 + \frac{f(xy)}{f(x)} = f(-x)^2 + \frac{f(-xy)}{f(-x)} = f(x)^2 + \frac{f(- xy)}{f(-x)} $$ hence taking $x = 1$ $$ \frac{f(y)}{f(-y)} = \frac{f(1)}{f(-1)}. $$ Thus depending on the sign of the rhs, $f$ is either even or odd.

If $f$ is even then $$f(x^2 + y) = f(x)^2 + \frac{f(xy)}{f(x)} = f(x)^2 + \frac{f(-xy)}{f(x)} = f(x^2 - y).$$ Now for any $z > 1$, let $x^2 = (z + 1)/2$ and $y = (z - 1)/2$. We get that $f(z) = f(1)$. In particular $f(1) = f(2) = f(1^2 + 1) = f(1)^2 + 1$ so $0 = f(1)^2 - f(1) + 1$. Taking the discriminant we see $f(1)$ is not real, contradiction. So $f$ is odd.

Now $$f(x^2 - y) = f(x)^2 + \frac{f(-xy)}{f(x)} = f(x)^2 - \frac{f(xy)}{f(x)}$$ so in particular $f(x^2 - 1) = f(x)^2 - 1$. Observe now that since $$ f(x^2 + 1) = f(x)^2 + 1 $$ we know $f(x) > 1$ for $x > 1$. Furthermore, for $0 < z < 1$ set $z = x^2 - 1$ so $x = \sqrt{z + 1} > 1$, hence we know $f(z) = f(x^2 - 1) = f(x)^2 - 1 > 0$. So we have shown that $f$ is positive-valued on $x > 0$. Hence (being lazy) I now appeal to the RELATED PROBLEM to answer the question. Now the intended solution is definitely not as complicated as this (already we know a lot more than in the RELATED PROBLEM), so I encourage you to find your own solution!

BELOW RELATED PROBLEM: as pointed out, below is a solution for $f : \mathbb{R}^{>0} \to \mathbb{R}^{>0}$, with the restriction $x,y > 0$ in the functional equation which is not the original problem. Actually it is probably harder because $y > 0$ is an annoying constraint.

Because $f$ is positive valued, we know that $$ f(x^2 + y) > f(x)^2 \tag{1} $$ and $$f(x^2 + y) > \frac{f(xy)}{f(x)} \tag{2}. $$ The following two Observations motivate us to compute $f(1)$ somewhat "analytically":

1. Taking $y = 1$ gives $f(x^2 + 1) > 1$, i.e. $f(x) > 1$ for $x > 1$.
2. The solutions to $x^2 + y = x$ are of the form $(x,x - x^2)$ where $x < 1$. Then from (1) we conclude that for any $x < 1$, $f(x) > f(x)^2$ i.e. $f(x) < 1$.

Let's repeat the latter trick, plugging into the actual equation. Then for $x < 1$ we have $$f(x) = f(x)^2 + \frac{f(x^2 - x^3)}{f(x)}$$ and rearranging, $$ f(x^2 - x^3) = f(x)^2 - f(x)^3. $$

Let $u_0 = 1/2$ (or an arbitrary value below $1$) and inductively let $u_{i + 1} = u_i^2 - u_i^3$. Because $u_{i + 1} < u_i^2 < 1$ and $u_{i + 1} = u_i^2 - u_i^3 > 0$ we have that $\lim_{i \to \infty} u_i = 0$, and a similar argument gives that $\lim_{i \to \infty} f(u_i) = 0$. Now $$ f(1 + u_i) = f(1)^2 + \frac{f(u_i)}{f(1)} $$ and taking the limit $$ \lim_{i \to \infty} f(1 + u_i) = f(1)^2. $$ Because $1 < f(1 + u_i)$ by Observation 1 we get that $1 \le f(1)^2$ hence $1 \le f(1)$.

Now go back to the equality and plug in $y = 1$. We get $$f(x^2 + 1) = f(x)^2 + 1$$ and taking $x = u_i$ and a limit $$\lim_{i \to \infty} f(u_i^2 + 1) = 1. $$ Taking $x = 1$, $y = u_i^2$ in the equality $$ f(1 + u_i^2) = f(1)^2 + \frac{f(u_i^2)}{f(1)} $$ and taking the limit we get $$ 1 = \lim_{i \to \infty} \left(f(1)^2 + \frac{f(u_i^2)}{f(1)}\right) $$ so in particular $f(1)^2 \le 1$, i.e. $f(1) \le 1$.

So finally we have shown $f(1) = 1$. Taking $x = 1$ $$ f(1 + y) = f(1)^2 + \frac{f(y)}{f(1)} = 1 + f(y) \tag{3}$$ so by induction $$f(n + r) = n + f(r)$$ for $n \in \mathbb{Z}^{> 0}$, $r \ge 0$. Taking $y = 1$ in the original equation $$ f(x^2 + 1) = f(x)^2 + 1 $$ so by (3), for all $x$ $f(x^2) = f(x)^2$, i.e. $f(\sqrt{x}) = \sqrt{f(x)}$. Now recalling the original equation $$ f(x^2 + y) = f(x)^2 + \frac{f(xy)}{f(x)} = f(x^2) + \frac{f(xy)}{f(x)} \tag{4} $$ so $f$ is an increasing function. Recalling our sequence $u_i$ with $u_i \to 0, f(u_i) \to 0$ as $i \to \infty$, we conclude that $\lim_{x \to^+ 0} f(x) = 0$. Taking the limit in (4), $$ \lim_{y \to^+ 0} f(x^2 + y) = f(x^2) $$ and so $f$ is right-continuous everywhere. Fix arbitrary $z$, and let $x^2 = z - y$. Then $$ f(z) - f(z - y) = f(x^2 + y) - f(x^2) = \frac{f(xy)}{f(x)} < \frac{f(\sqrt{z} \cdot y)}{f(\sqrt{z - k})} $$ for any fixed $k > 0$ such that $0 < z - k$, for all $y < k$ (so $\sqrt{z - k} < \sqrt{z - y}$). Taking the limit as $y \to^+ 0$ we find that $f$ is left-continuous at $z$. Hence $f$ is continuous everywhere.

I now use the (nontrivial, but intuitive) fact that non-negative multiples of $\sqrt{2}$ are dense in $\mathbb{R}/\mathbb{Z}$. In other words $$ \{a\sqrt{2} + b : a \in \mathbb{Z}^{\ge 0},b \in \mathbb{Z}\} $$ is dense in $\mathbb{R}$. For any positive $a\sqrt{2} + b = \sqrt{2a^2} + b$, if $b \ge 0$ then we know $$ f(\sqrt{2a^2} + b) = b + f(\sqrt{2a^2}) = b + \sqrt{2a^2}. $$ If $b < 0$ then we know $$ f(\sqrt{2a^2} + b) - b = f(\sqrt{2a^2} + b - b) = \sqrt{2a^2}. $$ So $f$ equals the identity on a dense subset of $\mathbb{R}^{> 0}$. By continuity, $f = id$.

PS This might not be the shortest solution! At least it is fairly systematic. Also, my apologies, there may be small mistakes, so caveat emptor.

Answer 3

$\newcommand{\NN}{{\mathbb{N}}}\newcommand{\CC}{{\mathbb{C}}}\newcommand{\RR}{{\mathbb{R}}}\newcommand{\QQ}{{\mathbb Q}}\newcommand{\ra}{\rightarrow}\newcommand{\ds}{\displaystyle}\newcommand{\mat}[1]{\left(\begin{matrix}#1\end{matrix}\right)}\newcommand{\lam}{\lambda}\renewcommand{\P}{{\mathcal P}}\newcommand{\N}{{\mathcal N}}\newcommand{\M}{{\mathcal M}}\renewcommand{\L}{{\mathcal L}}\newcommand{\EQ}[2]{\begin{equation} {{#2}\label{#1}} \end{equation}}\newcommand{\eps}{\varepsilon}$ It contains numerous elements that are also in other answers - I put it anyway. Observe that $f$ cannot be constant, because the equation $C=C^2+1$ has no real solution.

Putting $y=1$, we find that $f(x)^2+1=f(x^2+1)=f(-x)^2+1$ and hence $f(x)^2=f(-x)^2$ for all $x\in A$. Putting $x=\pm1$, we obtain $f(-y)/f(-1)=f(y)/f(1)$ for all $y$. So either $f(-x)=f(x)$ for all $x$ or $f(-x)=-f(x)$ for all $x$.

The solution $y$ of $x^2+y=xy$ is $y=\frac{x^2}{x-1}$ unless $x=1$. Using this $y$ we have $$f\left(\frac{x^3}{x-1}\right)=\frac{f(x)^3}{f(x)-1}\mbox{ for }x\neq1.$$ Consider now a solution $z$ of $\frac{z^2}{z-1}=-1$. Then $f(-z)=\frac{f(z)^3}{f(z)-1}$. As a consequence $f(-z)$ cannot equal $f(z)$ because $1=\frac{f(z)^2}{f(z)-1}$ has no real solution for $f(z)$. Hence $f(-z)=-f(z)$ and, as seen above, $f(-x)=-f(x)$ for all $x.$

Next consider $x=1$ and replace $y$ by $-y$. We find $f(1-y)=f(1)^2-\frac{f(y)}{f(1)}$ for $y\neq1$. Replacing here $y$ by $1-y$ yields with $w=f(1)$ $$w^2f(y)=w^4-wf(1-y)=w^4-w^3+f(y)\mbox{ for all }y\in A,y\neq1.$$ Thus $(w^2-1)f(y)=w^4-w^3$. So we must have $w=1$ as otherwise $w=-1$ which leads to a contradiction or $f(y)=\frac{w^3}{w+1}$ is constant which had been excluded. So finally we proved that $w=f(1)=1$.

Putting $x=1$ now gives $f(y+1)=f(y)+1$ for $y\neq-1$. In particular, we find $f(n)=n$ and $f(-n)=-n$ for all positive integers $n$. Putting $x=n$, we obtain $f(y+n^2)=n^2+\frac1nf(ny)$ for all $y\neq-n^2$ and $f(ny)=nf(y)$ for all positive $y$ (In fact, we only have to exclude that $y$ is a negative integer). In particular this leads to $f(q)=q$ for all rational positive $q$. It is extended to negative rationals using $f(-x)=-f(x)$. Moreover, we obtain $f(y+m)=f(y)+m$ for all non-integer $y$ and thus $f(y+\frac mn)=\frac1nf(ny+m)=\frac1n(nf(y)+m)=f(y)+\frac mn$ for all irrational $y$ and hence $f(y+q)=f(y)+q$ for all rational $q$ and irrational $y$.

Finally, as $f(x^2+1)=f(x)^2+1$ and hence $f(x^2)=f(x)^2$ for all $x$, the values of $f$ for positive $y$ are positive and hence also its values for negative $y$ are negative. Given any irrational $y$, we consider rational $q,r$ with $q<y<r$. Then as $y-q>0$, we have $f(y)-q=f(y-q)>0$ and hence $q<f(y)$. In the same way we show that $f(y)<r$. Since we can use any rational $q,r$ with $q<y<r$, we have shown that $f(y)=y$ also for irrational $y$.