Some people use the term one-to-one for injective, onto for surjective, and one-to-one onto for bijective functions.

A function \(f\) has an inverse if and only if it is bijective.

A strictly increasing or decreasing function is always injective.

The phrase if and only if cannot be stressed enough! If you think you’ve found all the solutions, chances are you probably haven’t. Many points are taken off because of the lack of the proof for the if and only if condition, and it is always nice to suspect your assumption that you have found all the solutions.

Solution.

Unlike the traditional approach of substituting different values to \(x\) and \(y\), let’s start by substituting \(y = 1\). Substituting, \(f(f(x) + f(1)) = x + 1 = f(f(x + 1) + f(0))\). To find injectivity, surjectivity, or bijectivity of the function \(f\), assume \(f(a) = f(b)\). Then, \(f(f(a) + f(y)) = f(f(b) + f(y))\) and by definition, \(f(a) = f(b)\) leads to \(a = b\). In other words, \(f\) is injective.

Returning to our previous equation \(f(f(x) + f(1)) = f(f(x + 1) + f(0))\), we could see that \(f(x) + f(1) = f(x + 1) + f(0)\) since \(f\) is injective. Rearranging, \(f(x + 1) - f(x) = f(1) - f(0)\) and \(f(x)\) is arithmetic sequence. Therefore let \(f(x) = ax + b\). Substituting to our given equation, \(f(ax + b + ay + b) = a (ax + b + ay + b) + b = x + y\). Solving the equation, \begin{align*} a (ax + b + ay + b) + b &= x + y \\ \left ( a^2 - 1 \right ) x + \left ( a^2 - 1 \right ) y + 2ab + b &= 0 \end{align*}

holds for all \(x, y \in \mathbb {R}\). Thus \(a = \pm 1\) and \(b = 0\). Therefore, \(f(x) = \pm x\) are all solutions.

Solution.

First, let’s start with what we can observe from the problem. The first thing that we can see is that we have squares everywhere. The second thing to note is that substituting \(a = b = c = d = 1\) gives us \(f(x) = 1\). We can also notice that if we substitute \((n, n, n, n)\) we get \(f(n)^2 = f(n^2)\).

With the observations in mind, what we can try is getting rid of the square by substituting \((\sqrt {x}, \sqrt {x}, 1, x)\). This will lead us to following equations. \begin{align*} \frac { f(\sqrt {x})^2 + f(\sqrt {x})^2 }{ 1 + f(x^2) } &= \frac {x + x}{1 + x^2} \\ \frac { 2f(x) }{ 1 + f(x)^2 } &= \frac {2x}{1 + x^2} \\ 2f(x) (1 + x^2) &= 2x(1 + f(x)^2) \end{align*}

Continuing, we can rearrange the equations to solve for \(f\). \begin{align*} xf(x)^2 - (1 + x^2)f(x) + x &= 0 \\ (xf(x) - 1) (f(x) - x) &= 0 \\ \therefore f(x) = x \text { or } f(x) &= \frac {1}{x} \end{align*}

Now we can see that if a function \(f\) satisfy the equation given by the problem, then \(f(x) = x\) or \(f(x) = \frac {1}{x}\). Moreover, if \(f(x) = x)\) or \(f(x) = \frac {1}{x}\), then \(f\) satisfies the original equation as we can see from direct substitution.

If we end our solution here thinking that we have solved the problem, we will lose points for not showing that \(x\) and \(\frac {1}{x}\) are the only solutions. Although we showed the if and only if condition, we have not showed that \(f\) is not a piece-wise function.

Proof. For the sake of contradiction, let \(f(\alpha ) \neq \alpha \) and \(f(\beta ) \neq \frac {1}{\beta }\) for \(\alpha , \beta \neq 1\). By our previous conclusions, we can see that it implies \(f(\alpha ) = \frac {1}{\alpha }\) and \(f(\beta ) = \beta \).

Substitute \((\sqrt {\alpha }, \sqrt {\beta }, 1, \sqrt {\alpha \beta })\) to eliminate the squares and use both \(\alpha \) and \(\beta \). \begin{align*} \frac { f(\sqrt {\alpha })^2 + f(\sqrt {\beta })^2 } { f(1) + f(\alpha \beta ) } &= \frac {\alpha + \beta }{1 + \alpha \beta } \\ \frac { \frac {1}{\alpha } + \beta }{ 1 + f(\alpha \beta ) } &= \frac {\alpha + \beta }{1 + \alpha \beta } \end{align*}

Notice that \(f(\alpha \beta )\) is either \(\alpha \beta \) or \(\frac {1}{\alpha \beta }\). Therefore, let \(f(\alpha \beta ) = \alpha \beta \). Solving the equation, the following equations are obtained. \begin{align*} \frac { \frac {1}{\alpha } + \beta }{ 1 + f(\alpha \beta ) } &= \frac {\alpha + \beta }{1 + \alpha \beta } \\ \frac { \frac {1}{\alpha } + \beta }{ 1 + \alpha \beta } &= \frac {\alpha + \beta }{1 + \alpha \beta } \\ \frac {1}{\alpha } &= \alpha \end{align*}

This contradicts with the fact that \(\alpha \neq 1\). Continuing, let \(f(\alpha \beta ) = \frac {1}{\alpha \beta }\). \begin{align*} \frac { \frac {1}{\alpha } + \beta }{ 1 + f(\alpha \beta ) } \displaybreak &= \frac {\alpha + \beta }{1 + \alpha \beta } \\ \frac { \frac {1}{\alpha } + \beta }{ 1 + \frac {1}{\alpha \beta } } &= \frac {\alpha + \beta }{1 + \alpha \beta } \\ \frac {\beta + \alpha \beta ^2}{\alpha \beta + 1} &= \frac {\alpha + \beta }{1 + \alpha \beta } \\ \alpha \beta ^2 &= \alpha \\ \beta ^2 &= 1 \end{align*}

This also contradicts with \(\beta \neq 1\). Therefore, there exists no \(\alpha , \beta \neq 1\) such that \(f(\alpha ) \neq \alpha \) and \(f(\beta ) \neq \frac {1}{\beta }\). □

Hence, \(f(x) = x\) and \(f(x) = \frac {1}{x}\) are the only solutions.

Solution.

First, let’s start by doing typical substitutions. Substituting \(x = y = 0\) gives \(f(f(0)) = f(0)\) and \(x = -y\) gives \(f(f(x)) = f(0) - xf(-x)\). Also, trying \(y = 0\) gives \(f(f(x) + x) = f(x)\). Looking at the last observation, we could assume that \(f(x) = 0\). Moreover our first observation also suggests that \(f(0) = 0\). With the observations, let’s first investigate the value of \(f(0)\).

Consider the following equation by the last observation with \(y = f(x) + x\). \begin{align*} f(x) &= f(y) = f(f(y) + y) = f(f(x) + f(x) + x) \\ &= f(x + f(x)) + f(x) f(f(x)) = f(x) + f(x) f(f(x)) \end{align*}

Therefore, \(f(x) f(f(x)) = 0\) and \(f(0) f(f(0)) = 0\). Note that because \(f(f(0)) = f(0)\), \(f(0)^2 = 0\) and \(f(0) = 0\) holds. Indeed \(f(0) = 0\)!

Continuing, substitute \(x = 0\) to the given equation. \[ f(f(0) + 0 + y) = f(y) = f(0 + y) + y f(y) \] In other words, \(y f(y) = 0\) holds for all \(y \in \mathbb {R}\). Specially, \(f(y) = 0\) for all \(y \neq 0\) and we have found that \(f(0) = 0\). Therefore, \(f(x) = 0\) for all \(x \in \mathbb {R}\) is the only solution.