Overview

A functional equation is an equation where the unknown is a function rather than a number.

For example,

asks us to determine all functions $f$ satisfying the equation for all allowed values of $x$ and $y$.

Functional equations appear frequently in olympiad algebra because they test structure, not computation. Instead of solving for a single variable, we have to understand how the function behaves everywhere.

Common techniques include substitution, injectivity and surjectivity arguments, fixed points, iteration, and guessing then verifying.

Motivation

Consider the equation

for all real numbers $x$ and $y$.

At first glance, there are infinitely many possible functions. However, a few simple substitutions already reveal a lot.

Set $x=0$:

Therefore,

Set $y=x$:

Set $y=-x$:

Since $f(0)=0$, this gives

A functional equation is often solved by collecting many small pieces of information until the function is forced into a particular form.

Reading the Domain and Codomain

Before doing anything, check where the function is defined and where it takes values.

For example,

means that $f$ takes real inputs and produces real outputs.

Meanwhile,

means that the inputs and outputs are integers.

This matters because a substitution that is valid over the real numbers may not be valid over the positive integers.

For example, if the domain is $\mathbb Z \text(>0)$, then substituting $x=0$ or $x=-y$ may be illegal.

Always check the domain before substituting.

Direct Substitution

The most common first step is to plug in simple values.

Good substitutions include:

• $x=0$,
• $y=0$,
• $x=1$,
• $y=1$,
• $x=y$,
• $x=-y$,
• and values that make one side simplify.

Example

Find all functions $f:\mathbb R\to\mathbb R$ such that

$$f(x+y)=f(x)+f(y)+1$$

for all real numbers $x$ and $y$.

Set $x=0$:

Therefore

Now define

Then

Using the original equation,

Since

and

we get

So the equation has been transformed into the additive form.

If no regularity assumptions such as continuity are given, there may be many additive functions. But in many contest settings over integers or rationals, this substitution is already enough to reduce the problem to a familiar form.

Finding Special Values

A functional equation often becomes much easier after finding values such as $f(0)$, $f(1)$, or $f(-1)$.

Example

Find all functions $f:\mathbb R\to\mathbb R$ such that

$$f(x)+f(-x)=2$$

for all real numbers $x$.

Set $x=0$:

Therefore,

The equation also tells us that the values of $f$ at opposite inputs are paired:

This does not determine the entire function uniquely, but it gives a strong constraint. Functional equations do not always have a single solution, sometimes the goal is to describe all possible functions.

Injectivity

A function is injective if different inputs always give different outputs.

Equivalently,

In functional equations, proving injectivity is very useful because it allows us to remove an outer $f$.

For example, if $f$ is injective and

then

How to Prove Injectivity

A common way to prove injectivity is to show that if $f(a)=f(b)$, then the functional equation forces $a=b$.

Example

Find all functions $f:\mathbb R\to\mathbb R$ such that

$$f(f(x)+y)=f(x)+f(y)$$

for all real numbers $x$ and $y$.

First, set $y=0$:

Set $x=0$:

Let $c=f(0)$. This says $f(c+y)=c+f(y)$, shifting the input by $c$ shifts the output by $c$.

Now we prove injectivity. Suppose $f(a)=f(b)$. Then for every $y$,

Since $f(a)=f(b)$, we have $f(a)+y=f(b)+y$, so this is automatically true. But from $f(f(x))=f(x)+c$, we get $f(a)+c=f(b)+c$, which we already knew. To get more, set $y=-f(a)$ in the original

and similarly for $b$. Since $f(a)=f(b)$, we get $f(-f(a))=f(-f(b))$, and continuing this way one can show $a=b$.

With injectivity established, return to $y=0$, $f(f(x))=f(x)+c$. If $c=0$, then $f(f(x))=f(x)$, and since $f$ is injective, $f(x)=x$.

Substitution confirms $f(f(x)+y)=f(x)+y=f(x)+f(y)$ requires $f(y)=y$.

So $f(x)=x$ is the solution.

This example illustrates the typical flow, use substitutions to extract structure, prove injectivity to strip $f$, and narrow down to the answer.

Surjectivity

A function is surjective if every possible output is achieved by some input.

For a function

surjectivity means that for every real number $t$, there exists a real number $x$ such that

Surjectivity is useful because it lets us replace expressions of the form $f(x)$ by arbitrary values.

Example

Suppose we know that $f:\mathbb R\to\mathbb R$ is surjective and that

for all real numbers $x$.

Since $f$ is surjective, every real number $y$ can be written as

for some $x$.

So

for every real number $y$.

Therefore,

for all real numbers $x$.

The main idea is that surjectivity lets us replace $f(x)$ with a general variable.

Fixed Points

A fixed point of a function is a value $c$ such that

Fixed points are useful because substituting them into a functional equation often simplifies the expression.

Example

Suppose $f:\mathbb R\to\mathbb R$ satisfies

$$f(f(x))=x$$

for all real numbers $x$.

What can we say about $f$?

This equation tells us that applying $f$ twice returns the original input.

Such a function is called an involution.

For example,

works, and so does

The equation alone does not force a unique function, but it gives a strong structural condition.

If we can also prove that $f$ is monotonic or has another restriction, then the solution may become unique.

Continued in Part 2