IMO 2002 P5

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Problem 1. Determine all functions

𝑓 ​ : ℝ ​ \to ℝ

such that

(𝑓 (𝑥) ​ + 𝑓 (𝑦)) (𝑓 (𝑢) ​ + 𝑓 (𝑣)) ​ = 𝑓 (𝑥 𝑢 ​ - 𝑦 𝑣) ​ + 𝑓 (𝑥 𝑣 ​ + 𝑦 𝑢)

Solution. Let $𝑥 = 𝑦 = 𝑢 = 𝑣 = 0$ to obtain $4 𝑓 (0) 2 = 2 𝑓 (0)$ , which implies $𝑓 (0) = 0$ or $12$ $12$ . We consider the problem in two cases.

$𝑓 (0) = 0$ .
Let $𝑢 = 𝑣 = 1$ to obtain $2 𝑓 (1) 𝑓 (𝑥) + 2 𝑓 (1) 𝑓 (𝑦) = 𝑓 (𝑥 - 𝑦) + 𝑓 (𝑥 + 𝑦)$ . Let $𝑥 = 0$ , then $2 𝑓 (1) 𝑓 (𝑦) = 𝑓 (𝑦) + 𝑓 ( - 𝑦)$ . Hence, $(2 𝑓 (1) - 1) 𝑓 (𝑦) = 𝑓 ( - 𝑦)$ . Let $𝑥 = 1, 𝑢 = 1, 𝑦 = 𝑣 = 0$ to obtain $𝑓 (1) 2 = 𝑓 (1)$ .
1. $𝑓 (1) = 0$ . Then $𝑓 (𝑥 + 𝑦) + 𝑓 (𝑥 - 𝑦) = 0$ . Let $𝑥 = 𝑦$ to obtain $𝑓 (2 𝑥) = 0$ . Hence, $𝑓 (𝑥) \equiv 0$ .
2. $𝑓 (1) = 1$ . There are two approaches. I state this case as a lemma and provide two proofs. See them below.
$𝑓 (0) = 12$ .
Let $𝑢 = 𝑣 = 1$ to obtain $2 𝑓 (1) 𝑓 (𝑥) + 2 𝑓 (1) 𝑓 (𝑦) = 𝑓 (𝑥 - 𝑦) + 𝑓 (𝑥 + 𝑦)$ . Let $𝑥 = 0$ , then $𝑓 (1) + 2 𝑓 (1) 𝑓 (𝑦) = 𝑓 ( - 𝑦) + 𝑓 (𝑦)$ . When $𝑦 = 0$ , it follows that $2 𝑓 (1) = 1$ . Hence, $𝑓 (1) = 12$ . Thus, $𝑓 (𝑦) + 𝑓 ( - 𝑦) = 12 + 𝑓 (𝑦)$ . Therefore $𝑓 (𝑥) \equiv 12$ .

Therefore, the final answers are $𝑓 (𝑥) \equiv 0$ , $𝑓 (𝑥) \equiv 12$ and $𝑓 (𝑥) = 𝑥 2$ .

Lemma 2. Given function $𝑓 : ℝ \to ℝ$ with
$(𝑓 (𝑥) + 𝑓 (𝑦)) (𝑓 (𝑢) + 𝑓 (𝑣)) = 𝑓 (𝑥 𝑢 - 𝑦 𝑣) + 𝑓 (𝑥 𝑣 + 𝑦 𝑢)$
$𝑓 (0) = 0$
$𝑓 (1) = 1$
Prove that $𝑓 (𝑥) = 𝑥 2$ .

Proof. Take

𝑢 ​ = 𝑣 ​ = 1, 𝑥 ​ = 0

to obtain

2 𝑓 (𝑦) ​ = 𝑓 (𝑦) ​ + 𝑓 (​ - 𝑦)

, which implies

𝑓 (𝑦) ​ = 𝑓 (​ - 𝑦)

. Let

𝑦 ​ = 𝑣 ​ = 0

, then

𝑓 (𝑥) 𝑓 (𝑢) ​ = 𝑓 (𝑥 𝑢)

. Hence,

𝑓 (𝑥 2) ​ = 𝑓 (𝑥) 2 ​ \geq 0

, which implies

𝑥 ​ \geq 0 ​ \Rightarrow 𝑓 (𝑥) ​ \geq 0

. Let

𝑢 ​ = 𝑥

𝑣 ​ = ​ - 𝑦

to obtain

(𝑓 (𝑥) + 𝑓 (𝑦)) 2 ​ = 𝑓 (𝑥 2 ​ + 𝑦 2)

. Let

𝜑 (𝑥) ​ = 𝑓 (\sqrt 𝑥), 𝑥 ​ \in [0, ​ + \infty)

, then by multiplication, we have

𝜑 (𝑥) ​ = \sqrt 𝑓 (𝑥) ​ \geq 0

and

𝜑 (𝑥) 𝜑 (𝑦) ​ = 𝜑 (𝑥 𝑦)

. Hence,

𝜑 (𝑥 2) ​ + 𝜑 (𝑦 2) ​ = 𝜑 (𝑥 2 ​ + 𝑦 2)

. Since

𝜑 (𝑥) ​ \geq 0 ​ \forall 𝑥 ​ \in [0, ​ + \infty)

, we obtain

𝜑

𝜑

to be monotone on its domain. Hence,

\forall 𝑥 ​ \geq 0, 𝜑 (𝑥) ​ = 𝑘 𝑥

, where

𝑘 ​ \in ℝ

. Since

𝑓 (1) ​ = 1

, it follows that

𝑘 ​ = 1

. Thus,

\forall 𝑥 ​ \geq 0, 𝜑 (𝑥) ​ = 𝑥

. Therefore

\forall 𝑥 ​ \geq 0, 𝑓 (𝑥) ​ = 𝑥 2

. By

𝑓 (𝑦) ​ = 𝑓 (​ - 𝑦)

, we obtain

𝑓 (𝑥) ​ = 𝑥 2

□

Proof. Let

𝑢 ​ = 𝑣 ​ = 1

, then

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

. Let

𝑥 ​ = 0

, it follows that

𝑓 (𝑦) ​ = 𝑓 (​ - 𝑦)

. Take

𝑦 ​ = 1

to obtain

𝑓 (𝑥 ​ + 1) ​ = 2 𝑓 (𝑥) ​ + 2 ​ - 𝑓 (𝑥 ​ - 1)

. Since

𝑓 (0) ​ = 0

𝑓 (1) ​ = 1

, by induction,

𝑓 (𝑛 ​ + 1) ​ = 2 𝑓 (𝑛) ​ + 2 ​ - 𝑓 (𝑛 ​ - 1) ​ = 2 𝑛 2 ​ + 2 ​ - (𝑛 - 1) 2 ​ = (𝑛 + 1) 2

implies

𝑓 (𝑛) ​ = 𝑛 2 ​ \forall 𝑛 ​ \in ℕ

. By

𝑓 (𝑦) ​ = 𝑓 (​ - 𝑦)

, it follows that

𝑓 (𝑛) ​ = 𝑛 2 ​ \forall 𝑛 ​ \in ℤ

. Let

𝑦 ​ = 𝑣 ​ = 0

to obtain

𝑓 (𝑥) 𝑓 (𝑢) ​ = 𝑓 (𝑥 𝑢)

, then it is not hard to show that

𝑓 (𝑞) ​ = 𝑞 2 ​ \forall 𝑞 ​ \in ℚ

. Since

𝑓 (𝑥) 𝑓 (𝑢) ​ = 𝑓 (𝑥 𝑢)

𝑓 (𝑥 2) ​ = 𝑓 (𝑥) 2 ​ \geq 0

, which implies

𝑥 ​ \geq 0 ​ \Rightarrow 𝑓 (𝑥) ​ \geq 0

. Let

𝑥, 𝑦 ​ \geq 0

𝑢 ​ = 𝑥

𝑣 ​ = ​ - 𝑦

to obtain

𝑓 (𝑥 2 ​ + 𝑦 2) ​ = (𝑓 (𝑥) + 𝑓 (𝑦)) 2 ​ = 𝑓 (𝑥) 2 ​ + 2 𝑓 (𝑥) 𝑓 (𝑦) ​ + 𝑓 (𝑦) 2 ​ = 𝑓 (𝑥 2) ​ + 2 𝑓 (𝑥) 𝑓 (𝑦) ​ + 𝑓 (𝑦) 2 ​ \geq 𝑓 (𝑥 2)

. Hence,

𝑓

𝑓

is monotone on

[0, ​ + \infty)

. Hence,

\forall 𝑥 ​ \geq 0, 𝑓 (𝑥) ​ = 𝑥 2

. By

𝑓 (𝑥) ​ = 𝑓 (​ - 𝑥)

, it follows immediately that

𝑓 (𝑥) ​ = 𝑥 2

□

View Typst source

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#show: tylenotes.with(
  title: "IMO 2002 P5",
  date: "2026-10-31",
  tags: ("imo", "lean"),
  summary: "Two solutions for IMO 2002 P5",
)

#problem[
  Determine all functions $f: RR->RR$ such that
  $(f(x)+f(y))(f(u)+f(v))=f(x u - y v) + f(x v + y u)$.
]

#solution[
  Let $x=y=u=v=0$ to obtain
  $4 f(0)^2 = 2 f(0)$,
  which implies $f(0) = 0$ or $1/2$.
  We consider the problem in two cases.
  + $f(0) = 0$. \
    Let $u = v = 1$ to obtain $2f(1)f(x)+2f(1)f(y)=f(x-y)+f(x+y)$.
    Let $x = 0$, then $2f(1)f(y)=f(y) + f(-y)$.
    Hence, $(2f(1)-1)f(y) = f(-y)$.
    Let $x=1,u=1,y=v=0$ to obtain $f(1)^2=f(1)$.
    + $f(1) = 0$.
      Then $f(x+y)+f(x-y) = 0$.
      Let $x=y$ to obtain $f(2x)=0$. Hence, $f(x) eq.triple 0$.
    + $f(1) = 1$.
      There are two approaches.
      I state this case as a lemma and provide two proofs.
      See them below.
  + $f(0) = 1/2$. \
    Let $u=v=1$ to obtain $2f(1)f(x)+2f(1)f(y) = f(x-y)+f(x+y)$.
    Let $x = 0$, then $f(1)+2f(1)f(y) = f(-y)+f(y)$.
    When $y = 0$, it follows that
    $2f(1)=1$.
    Hence, $f(1)=1/2$.
    Thus, $f(y)+f(-y) = 1/2+f(y)$.
    Therefore $f(x) eq.triple 1/2$.
  Therefore, the final answers are
  $f(x) eq.triple 0$, $f(x) eq.triple 1/2$
  and $f(x) eq x^2$.
]

#lemma[
  Given function $f:RR->RR$ with
  + $(f(x)+f(y))(f(u)+f(v))=f(x u - y v) + f(x v + y u)$
  + $f(0) = 0$
  + $f(1) = 1$
  Prove that $f(x) = x^2$.
]

#proof[
  Take $u = v = 1, x = 0$ to obtain
  $2f(y) = f(y) + f(-y)$, which implies $f(y) = f(-y)$.
  Let $y = v = 0$, then $f(x)f(u)=f(x u)$.
  Hence, $f(x^2) = f(x)^2 >= 0$,
  which implies $x >= 0 => f(x) >= 0$.
  Let $u = x$, $v = -y$ to obtain
  $(f(x)+f(y))^2 = f(x^2+y^2)$.
  Let $phi(x) = f(sqrt(x)), x in [0, +infinity)$,
  then by multiplication,
  we have $phi(x) = sqrt(f(x)) >= 0$
  and $phi(x)phi(y) = phi(x y)$.
  Hence, $phi(x^2) + phi(y^2) = phi(x^2 + y^2)$.
  Since $phi(x) >= 0 thick forall x in [0,+infinity)$,
  we obtain $phi$ to be monotone on its domain.
  Hence, $forall x >= 0, phi(x) = k x$, where $k in RR$.
  Since $f(1) = 1$, it follows that $k = 1$.
  Thus, $forall x >= 0, phi(x) = x$.
  Therefore $forall x >= 0, f(x) = x^2$.
  By $f(y) = f(-y)$, we obtain $f(x) = x^2$.
]

#proof[
  Let $u = v = 1$,
  then $2f(x) + 2f(y) = f(x+y) + f(x-y)$.
  Let $x = 0$, it follows that $f(y) = f(-y)$.
  Take $y = 1$ to obtain
  $f(x+1) = 2f(x)+2-f(x-1)$.
  Since $f(0) = 0$, $f(1) = 1$,
  by induction,
  $f(n+1) = 2f(n)+2-f(n-1) = 2n^2+2-(n-1)^2=(n+1)^2$
  implies $f(n) = n^2 thick forall n in NN$.
  By $f(y) = f(-y)$, it follows that $f(n) = n^2 thick forall n in ZZ$.
  Let $y = v = 0$ to obtain $f(x)f(u)=f(x u)$,
  then it is not hard to show that
  $f(q) = q^2 thick forall q in QQ$.
  Since $f(x)f(u) = f(x u)$,
  $f(x^2) = f(x)^2 >= 0$,
  which implies $x>=0 => f(x) >= 0$.
  Let $x,y >= 0$, $u = x$, $v = -y$ to obtain
  $f(x^2+y^2) = (f(x)+f(y))^2 = f(x)^2 + 2f(x)f(y)+f(y)^2
  = f(x^2) + 2f(x)f(y)+f(y)^2 >= f(x^2)$.
  Hence, $f$ is monotone on $[0, +infinity)$.
  Hence, $forall x >= 0, f(x) = x^2$.
  By $f(x) = f(-x)$, it follows immediately that
  $f(x) = x^2$.
]