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Overview

Advanced functional equations often mix algebra and number theory. Expect to find hidden constraints via special inputs and symmetry.

Key Ideas

  • If ff is defined on integers, parity and modular arguments can restrict it.
  • Look for fixed points: solve f(x)=xf(x)=x or f(x)=cf(x)=c.
  • Iteration (apply the equation repeatedly) can uncover structure.

Core Skills

Pin Down Constants

Use x=0x=0, y=0y=0, x=yx=y, or x=−yx=-y to extract f(0)f(0) and other constants.

Iterate the Functional Equation

Apply the equation multiple times to relate f(x)f(x) to f(f(x))f(f(x)) or f(x+c)f(x+c).

Use Injectivity/Surjectivity

Prove injectivity by showing f(a)=f(b)⇒a=bf(a)=f(b)\Rightarrow a=b, then use it to simplify equations.

Worked Example

If f(x)=2xf(x)=2x satisfies f(x+y)=f(x)+f(y)f(x+y)=f(x)+f(y) for all x,yx,y, then ff is additive. Many contest problems ask you to prove that linear functions are the only solutions.

Easy Warm-up Examples

Example 1: Fixed Point

If f(f(x))=xf(f(x))=x, show ff is bijective.

Apply ff to both sides to get f(x)=f(f(f(x)))f(x)=f(f(f(x))), implying surjectivity and injectivity.

Example 2: Integer Restriction

If f:Z→Zf:\mathbb{Z}\to\mathbb{Z} and f(x+1)=f(x)+2f(x+1)=f(x)+2, then f(x)=2x+cf(x)=2x+c.

Example 3: Symmetry

If f(x)+f(1−x)=1f(x)+f(1-x)=1, then setting x=1−xx=1-x gives f(1/2)=1/2f(1/2)=1/2.

Example 4: Injectivity Practice

If

f(x+f(y))=f(x)+yf(x+f(y))=f(x)+y

for all real numbers x,yx,y, your goal is to show that ff is injective.

Suppose f(a)=f(b)f(a)=f(b). Then

f(x+f(a))=f(x)+a,f(x+f(a))=f(x)+a,

and

f(x+f(b))=f(x)+b.f(x+f(b))=f(x)+b.

Since f(a)=f(b)f(a)=f(b), we obtain a=ba=b.

Example 5: Surjectivity Practice

If

f(f(x))=f(0)+xf(f(x))=f(0)+x

for all real numbers xx, your job is to show that ff is surjective.

As xx varies over R\mathbb{R}, the right-hand side attains every real value. Hence f(f(x))f(f(x)) attains every real value, so ff is surjective.

Example 6: What c makes the function 0?

Suppose

f(x+f(y))=f(x)+yf(x+f(y))=f(x)+y

for all real numbers x,yx,y.

Since ff is surjective, choose cc such that f(c)=0f(c)=0. Then

f(x)=f(x+f(c))=f(x)+c,f(x)=f(x+f(c))=f(x)+c,

so c=0c=0.

Example 7: Finding Solutions

Your job is to find all linear solutions to the following functional equation:

f(x+f(y))=f(x)+2y.f(x+f(y))=f(x)+2y.

First, we let f(x)=ax+bf(x)=ax+b.Then, substituting gives

a(x+ay+b)+b=ax+2y+b.a(x+ay+b)+b=ax+2y+b.

Comparing coefficients yields

a2=2,ab=0.a^2=2,\qquad ab=0.

Since a≠0a\neq 0, we get b=0b=0, so

f(x)=2 xorf(x)=−2 x.f(x)=\sqrt{2}\,x \quad\text{or}\quad f(x)=-\sqrt{2}\,x.

Example 8: What is f(0)f(0)?

Suppose that

f(x+y)+f(x−y)=2f(x)+2f(y)f(x+y)+f(x-y)=2f(x)+2f(y)

for all real numbers x,yx,y.

Setting x=y=0x=y=0 gives

2f(0)=4f(0),2f(0)=4f(0),

hence

f(0)=0.f(0)=0.

Example 9: Find the Functional Relation

Suppose

f(x+y)+f(x−y)=2f(x)+2f(y)f(x+y)+f(x-y)=2f(x)+2f(y)

and f(0)=0f(0)=0.

Setting y=xy=x gives

f(2x)+f(0)=4f(x),f(2x)+f(0)=4f(x),

so

f(2x)=4f(x).f(2x)=4f(x).

Example 10: Iteration

If f(f(x))=f(0)+xf(f(x))=f(0)+x, show that f(f(f(f(x))))=2f(0)+xf(f(f(f(x))))=2f(0)+x.

Applying the given relation with xx replaced by f(f(x))f(f(x)) gives

f(f(f(f(x))))=f(0)+f(f(x)).f(f(f(f(x))))=f(0)+f(f(x)).

Using the original equation again,

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

Example 11: Constant Solutions

Your goal here is to find all constant solutions of

f(x+y)=f(x)f(y)−f(xy)+1.f(x+y)=f(x)f(y)-f(xy)+1.

Let f(x)≡cf(x)\equiv c. Then

c=c2−c+1.c=c^2-c+1.

Thus

(c−1)2=0,(c-1)^2=0,

so the only constant solution is

f(x)≡1.f(x)\equiv 1.

Example 12: What is f(0)f(0)?

If

f(x+y)=f(x)f(y)−f(xy)+1f(x+y)=f(x)f(y)-f(xy)+1

for all real numbers x,yx,y, show that either f(0)=1f(0)=1 or ff is constant.

Setting x=0x=0 gives

f(y)=f(0)f(y)−f(0)+1.f(y)=f(0)f(y)-f(0)+1.

Letting a=f(0)a=f(0), we obtain

(1−a)f(y)=1−a.(1-a)f(y)=1-a.

Thus either a=1a=1 or

f(y)≡1.f(y)\equiv 1.

Medium-Level Examples

Strategy Checklist

  • Plug in special values to find constants.
  • Iterate the equation to uncover structure.
  • Use injectivity or surjectivity when possible.

Practice Problems

StatusSourceProblem NameDifficultyTags
AIMEVery Hard
Show TagsFunctional Equations
AIMEVery Hard
Show TagsFunctional Equations, Substitution
CustomHard
Show TagsFunctional Equations, Substitution
CustomHard
Show TagsCauchy Equation, Functional Equations
CustomHard
Show TagsFunctional Equations, Iteration
CustomHard
Show TagsFunctional Equations, Involutions
CustomHard
Show TagsFunctional Equations, Logarithmic Transformation
Berkeley Math CircleHard
Show TagsFunctional Equations, Induction / smoothing
Berkeley Math Circle: Monthly Contest 7Hard
Show TagsFunctional Equations, Simple Equations
Berkeley Math CircleHard
Show TagsFunctional Equations, Quadratic functions
Berkeley Math CircleHard
Show TagsFunctional Equations, Polynomial operations
Berkeley Math Circle Monthly Contest 3Hard
Show TagsExistential quantifiers, Functional Equations
Berkeley Math Circle Monthly Contest 6Hard
Show TagsFunctional Equations, Recurrence relations
Berkeley Math CircleHard
Show TagsFunctional Equations
Berkeley Math Circle Monthly Contest 1Hard
Show TagsFunctional Equations, Induction / smoothing, Injectivity / surjectivity
Berkeley Math Circle: Monthly Contest 6Hard
Show TagsFunctional Equations, Polynomials, Recurrence relations
Berkeley Math Circle Monthly Contest 4Hard
Show TagsFunctional Equations, Induction / smoothing
Berkeley Math Circle Monthly Contest 8Hard
Show TagsFunctional Equations, Recurrence relations
Harvard-MIT Mathematics TournamentHard
Show TagsFunctional Equations
HMMT 1998Hard
Show TagsFunctional Equations
Harvard-MIT Math TournamentHard
Show TagsFunctional Equations
Berkeley Math Circle Monthly Contest #1Hard
Show TagsFunctional Equations
Harvard-MIT Math TournamentHard
Show TagsFunctional Equations
USA IMO 2003Hard
Show TagsFunctional Equations
Harvard-MIT Mathematics TournamentHard
Show TagsFunctional Equations, Linear and quadratic inequalities

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