Math Core

A functional equation describes a function by a property it satisfies, like $f(x + y) = f(x) + f(y)$ , instead of giving a formula. AIME functional equations are designed to be cracked by a handful of moves: plug in clever values, substitute to create a system, and look for structure like symmetry or cycles.

Move 1: plug in special values

Try $x = 0$ , $y = 0$ , $x = y$ , $y = -x$ , $x = 1$ , or $y = 1$ . Each substitution gives a new equation, often pinning down $f(0)$ or $f(1)$ first. Once you know one value, substitute it back to learn more.

For example, suppose $f(x)f(y) - f(xy) = x + y$ for all reals $x, y$ . With $x = y = 0$ : $f(0)^2 - f(0) = 0$ , so $f(0) = 0$ or $f(0) = 1$ . With $y = 0$ : $f(x)f(0) - f(0) = x$ . If $f(0) = 0$ this says $0 = x$ for every $x$ , which is false. So $f(0) = 1$ and then $f(x) - 1 = x$ , giving $f(x) = x + 1$ . Always check the answer in the original equation: $(x + 1)(y + 1) - (xy + 1) = x + y$ , as required.

Move 2: substitute to build a system

If the equation mixes $f(x)$ with $f$ of a transformed input, like $f\!\left(\dfrac1x\right)$ or $f(1 - x)$ , replace $x$ by that transformed input. When the transformation undoes itself ( $\dfrac{1}{1/x} = x$ , $1 - (1 - x) = x$ ), you get a second equation in the same two unknowns and can solve for $f(x)$ like an ordinary linear system.

Cycles of substitutions

If $g$ is a map with $g(g(x)) = x$ , then an equation involving $f(x)$ and $f(g(x))$ becomes a $2 \times 2$ linear system after replacing $x$ by $g(x)$ .

If instead $g(g(g(x))) = x$ (a 3-cycle, such as $g(x) = \dfrac{1}{1 - x}$ ), apply the equation at $x$ , $g(x)$ and $g(g(x))$ to get a $3 \times 3$ system in $f(x)$ , $f(g(x))$ , $f(g(g(x)))$ .

You can check that $g(x) = \dfrac{1}{1 - x}$ cycles: $g(g(x)) = \dfrac{x - 1}{x}$ and $g\!\left(\dfrac{x-1}{x}\right) = x$ .

Move 3: recursion on the integers

If the equation holds for integers, set $y = 1$ . Then $f(n + 1)$ is expressed through $f(n)$ , and you can sum up to get a formula. For example, $f(x + y) = f(x) + f(y)$ on the integers gives $f(n + 1) = f(n) + f(1)$ , so $f(n) = nf(1)$ .

Move 4: use symmetry to pair terms

If $f(x) + f(c - x)$ is constant, a sum of $f$ over inputs symmetric about $\dfrac c2$ can be paired up. This turns a sum of hundreds of terms into a count.

Common mistake

Plugging in values only gives necessary conditions. The function you find might fail the original equation for other inputs, or there might be several possible functions. Always verify your candidate in the original equation, and if the problem hints at more than one solution (for example, "the sum of all possible values"), look for every case.

Worked examples

Worked example: Building a system

A function satisfies $f(x) + 2f\!\left(\dfrac1x\right) = 3x$ for all $x \ne 0$ . Find $f(x)$ .

Replace $x$ by $\dfrac1x$ : $f\!\left(\dfrac1x\right) + 2f(x) = \dfrac3x$ . Multiply this by $2$ and subtract the original equation:

4f(x) - f(x) = \frac6x - 3x \implies f(x) = \frac2x - x.

Check: $\left(\dfrac2x - x\right) + 2\left(2x - \dfrac1x\right) = 3x$ , as required.

Worked example: A recursion on the integers

A function on the integers satisfies $f(x + y) = f(x) + f(y) + xy$ and $f(1) = 1$ . Find $f(10)$ .

With $y = 1$ : $f(n + 1) = f(n) + 1 + n$ . So $f(n) = f(1) + (2 + 3 + \dots + n) = \dfrac{n(n+1)}{2}$ for positive $n$ , and $f(10) = 55$ .

Worked example: Pairing

Let $f(x) = \dfrac{4^x}{4^x + 2}$ . Find $f\!\left(\dfrac{1}{11}\right) + f\!\left(\dfrac{2}{11}\right) + \dots + f\!\left(\dfrac{10}{11}\right)$ .

Let $u = 4^x$ . Then $4^{1 - x} = \dfrac4u$ , so

f(1 - x) = \frac{4/u}{4/u + 2} = \frac{4}{4 + 2u} = \frac{2}{u + 2}, \qquad f(x) + f(1 - x) = \frac{u + 2}{u + 2} = 1.

The ten terms form five pairs $\left(\dfrac{k}{11}, \dfrac{11 - k}{11}\right)$ , so the sum is $5$ .

Worked example: Composition on the positive integers

A strictly increasing function $f$ from the positive integers to the positive integers satisfies $f(f(n)) = 3n$ . Find $f(5)$ .

If $f(1) = 1$ , then $f(f(1)) = 1 \ne 3$ . So $f(1) = a \ge 2$ , and $f(a) = 3$ . Since $f$ is increasing and $a > 1$ , $3 = f(a) > f(1) = a$ , so $a = 2$ . Then $f(2) = 3$ , $f(3) = f(f(2)) = 6$ and $f(6) = f(f(3)) = 9$ . Now $6 = f(3) < f(4) < f(5) < f(6) = 9$ leaves only $f(4) = 7$ and $f(5) = 8$ .

Tip

If a functional equation involves a polynomial $P$ , compare degrees first. For example, if $P(P(x))$ is linear, then $(\deg P)^2 = 1$ , so $P$ itself is linear.

Practice

Practice 1

A function $f$ on the positive integers satisfies $f(mn) = f(m) + f(n)$ for all positive integers $m, n$ . If $f(2) = 3$ and $f(3) = 5$ , find $f(72)$ .