Iteration

Some functional equations involve repeated application of the function, such as

or even longer chains.

This is called iteration.

Example

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

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

and

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

for all real numbers $x$.

The first equation says that applying $f$ twice increases the input by $2$.

If we guess that $f$ is linear, say

then

We need

As such

If $a=1$, then

Then

so $b=1$.

This gives

If $a=-1$, then

Then

which does not work.

So the natural linear solution is

Substitution confirms that it satisfies both equations.

Symmetry

Many functional equations contain two variables and remain similar after swapping them.

For example, if an equation holds for all $x$ and $y$, then it also holds after exchanging $x$ and $y$.

Example

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

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

and

$$f(xy)=f(x)f(y)$$

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

Set $y=0$ in the first equation:

so

Set $y=1$ in the second equation:

If there exists some $x$ with $f(x)\neq 0$, then

Now set $x=y=1$ in the first equation:

Also by repeated addition,

for all positive integers $n$.

This example shows how combining equations and simple substitutions can build information in increments.

A full classification over the real numbers requires more care, but contest problems often include enough restrictions to force the intended solution.

Guess and Verify

In many intermediate problems, the solution is a simple function such as:

or

It is often useful to guess a possible form, substitute it into the equation, and then prove that no other functions work.

Example

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

$$f(x+y)=f(x)+f(y)+3$$

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

Assume

Then

On the other hand,

Thus

So

There is no restriction on $a$, so all linear solutions are

Substitution confirms that all such functions work.

Worked Examples

Example 1

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

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

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

Set $x=0$:

Let

Then

So the only possible functions are

Substituting back:

and

They are equal, so every function

works.

Example 2

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

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

and $f(1)=3$, assuming $f$ is linear.

Since $f$ is linear, write

Set $x=0$ in the functional equation:

so

So

Since

we get

Therefore,

Substitution confirms that this function works.

Example 3

Find all functions $f:\mathbb R\to\mathbb R$ satisfying

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

and

$$f(x)+x=0$$

for all real numbers $x$.

The second equation gives

Now check the first equation:

So

works.

Example 4

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

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

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

Set $x=1$:

This tells us that $f$ takes every value of the form $y + f(1)$, which is every real number. So $f$ is surjective.

It also tells us that $f$ is injective, if $f(a)=f(b)$, then $f(a)+1=f(b)+1$, so $f(f(a)+1)=f(f(b)+1)$, giving $a+f(1)=b+f(1)$, hence $a=b$.

Now set $y=0$:

Since $f$ is injective, we can remove $f$ from both sides:

For all $x \neq 0$, this gives $f(0)=0$.

Set $x=1$ again, $f(f(y)+1)=y+f(1)$. Setting $y=0$: $f(f(0)+1)=f(1)$, so $f(1)=f(1)$, which is consistent.

Now guess $f(x)=x$. Then $f(xf(y)+x)=xy+x$ and $xy+f(x)=xy+x$. These match, so $f(x)=x$ is a solution.

To confirm uniqueness, from $x=1$ we have $f(f(y)+1)=y+f(1)$. Setting $y=-f(1)$ gives $f(f(-f(1))+1)=0$, so $f(-f(1))+1=0$ (since $f$ is injective and $f(0)=0$), meaning $f(-f(1))=-1$. Continuing with specific substitutions, one can show $f(1)=1$, then $f(n)=n$ for all integers, extend to rationals, and use monotonicity (which follows from the original equation) to conclude $f(x)=x$ for all real $x$.

This example combines multiple techniques, one substitution gives surjectivity and injectivity, another uses injectivity to remove $f$ and find $f(0)$, and guess and verify is used to complete it.

Key Contest Insight

Functional equations are usually solved by building information gradually.

A typical solution does not usually come from only one substitution. Instead, you collect facts such as:

• the value of $f(0)$,
• whether $f$ is injective,
• whether $f$ is surjective,
• how $f$ behaves at special inputs,
• and whether the equation suggests a linear or constant form.

Once a candidate function is found, always substitute it back into the original equation to verify that it works.

Strategy Tips

• Start with simple substitutions such as $0$, $1$, $x=y$, and $y=-x$.
• Check whether the domain allows your substitution.
• Try to find $f(0)$ and $f(1)$ early.
• If you see $f(A)=f(B)$, consider proving injectivity.
• If every value can be written as $f(x)$, use surjectivity.
• If the equation contains $f(f(x))$, think about iteration.
• If a simple function seems likely, guess it, test it, and then prove it is forced.
• Always verify all proposed solutions.

Common Pitfalls

• Substituting values that are not in the domain.
• Assuming the function is continuous or linear when this is not stated.
• Proving that a function is possible but forgetting to prove uniqueness.
• Forgetting to substitute the final answer back into the original equation.
• Cancelling or applying injectivity before it has been proved.
• Ignoring constant solutions.

Practice Problems