IMO 1992 P2

Mathematics/Problems/All notes31 Oct 2026

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Problem 1. Determine all functions 𝑓​:ℝ​→ℝ such that 𝑓(𝑥2​+𝑓(𝑦))​=𝑦​+𝑓(𝑥)2.

Note 2. We denote 𝑓2(𝑥)​=𝑓(𝑓(𝑥)).

Solution. Let 𝑥​=0, then 𝑓2(𝑦)​=𝑦​+𝑓(0)2. Hence, 𝑓2𝑓2 is bijective, which implies 𝑓𝑓 is bijective. Hence, take 𝑡​∈ℝ such that 𝑓(𝑡)​=0. Let 𝑎​=𝑓2(𝑡)​=𝑓(0). Let 𝑥​=𝑡, 𝑦​=0, then 𝑓(𝑡2​+𝑎)​=𝑓(𝑡)2​=0​=𝑓(𝑡). By bijectivity, we obtain 𝑡2​+𝑎​=𝑡. Let 𝑥​=0, 𝑦​=𝑡, it follows that 𝑎​=𝑓2(𝑡)​=𝑡​+𝑓(0)2​=𝑡​+𝑎2. Substitute 𝑡​=𝑡2​+𝑎 in to obtain 𝑡2​+𝑎2​=0. Since 𝑡2​≥0 and 𝑎2​≥0, it follows that 𝑡​=0. Thus, 𝑓(0)​=0. Let 𝑦​=0 to obtain 𝑓(𝑥2)​=𝑓(𝑥)2. Let 𝑎​=𝑥2,𝑏​=𝑓(𝑦), then 𝑎​∈[0,​+∞),𝑏​∈ℝ, and 𝑓(𝑎​+𝑏)​=𝑦​+𝑓(𝑥)2​=𝑓(𝑥2)​+𝑓(𝑓(𝑦))​=𝑓(𝑎)​+𝑓(𝑏). Then f is odd. Hence, 𝑓(𝑥​+𝑦)​=𝑓(𝑥)​+𝑓(𝑦)​∀𝑥,𝑦​∈ℝ. Since 𝑓(𝑥2)​=𝑓(𝑥)2​≥0, 𝑓(𝑥)𝑓(𝑥) is monotone, then 𝑓(𝑥)​=𝑘𝑥. Substitute this in to obtain 𝑘𝑥2​+𝑘2𝑦​=𝑦​+𝑘2𝑥2, then 𝑘2​−𝑘​=0 and 𝑘2​=1. Thus, 𝑘​=1, which implies 𝑓(𝑥)​=𝑥.

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  title: "IMO 1992 P2",
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#problem[
  Determine all functions $f: RR->RR$ such that
  $f(x^2+f(y))=y+f(x)^2$.
]

#note[
  We denote $f^2(x) = f(f(x))$.
]

#solution[
  Let $x = 0$, then $f^2(y)=y+f(0)^2$.
  Hence, $f^2$ is bijective,
  which implies $f$ is bijective.
  Hence, take $t in RR$ such that $f(t) = 0$.
  Let $a = f^2(t) = f(0)$.
  Let $x = t$, $y = 0$,
  then $f(t^2+a) = f(t)^2 = 0 = f(t)$.
  By bijectivity, we obtain $t^2 + a = t$.
  Let $x = 0$, $y = t$,
  it follows that
  $a = f^2(t) = t + f(0)^2 = t + a^2$.
  Substitute $t = t^2 + a$ in to obtain $t^2 + a^2 = 0$.
  Since $t^2 >= 0$ and $a^2 >= 0$, it follows that $t = 0$.
  Thus, $f(0) = 0$.
  Let $y = 0$ to obtain $f(x^2)=f(x)^2$.
  Let $a = x^2, b = f(y)$, then $a in [0,+infinity), b in RR$,
  and $f(a+b) = y + f(x)^2 = f(x^2) + f(f(y)) = f(a) + f(b)$.
  Then f is odd.
  Hence, $f(x+y)=f(x)+f(y) thick forall x,y in RR$.
  Since $f(x^2)=f(x)^2 >= 0$, $f(x)$ is monotone,
  then $f(x) = k x$.
  Substitute this in to obtain
  $k x^2 + k^2 y = y + k^2 x^2$,
  then $k^2-k=0$ and $k^2 = 1$.
  Thus, $k = 1$, which implies $f(x) = x$.
]