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Iranian Mathematical Olympiad

Algebra Difficulty 5.0 AIME, harder Prove it Iran

Let f,g:R+→R+f, g : \mathbb{R}^+ \to \mathbb{R}^+ be two functions such that for all positive real numbers xx and yy
f(x+g(y))2=f(x2)+y2. f(x + g(y))^2 = f(x^2) + y^2.
Prove that the range of gg is not bounded from above.

Solution

Let P(x,y)P(x, y) be the assertion
f(x+g(y))2=f(x2)+y2 f(x + g(y))^2 = f(x^2) + y^2
If g(f(t))<tg(f(t)) < t for some t>0t > 0, then P(t−g(f(t)),f(t))P(t - g(f(t)), f(t)) implies
f(t)2=f((t−g(t))2)+f(t)2  ⟹  f((t−g(t))2)=0 f(t)^2 = f((t - g(t))^2) + f(t)^2 \implies f((t - g(t))^2) = 0
which is impossible.
So g(f(x))≥xg(f(x)) \ge x, for all positive real numbers xx. Hence the solution is complete. ■

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