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Saudi Booklet

Algebra Difficulty 4.8 AIME Prove it Saudi Arabia

Find all functions f:R+→Rf : \mathbb{R}^+ \to \mathbb{R} such that for all x,y>0x, y > 0 we have
f(x2+xf(y)+y)=2x+f(y). f(x^2 + x f(y) + y) = 2x + f(y).

Solution

Let r>sr > s be two positive real numbers. Then, there would be a positive real number tt such that t2+f(s)t+s−r=0t^2 + f(s)t + s - r = 0. Now, put (x,y)=(t,s)(x, y) = (t, s) it follows that
f(r)=f(t2+f(s)t+s)=2t+f(s)>f(s). f(r) = f(t^2 + f(s)t + s) = 2t + f(s) > f(s).
So, ff is strictly increasing and hence injective. Plugging (x,y)=(f(z)2,y)(x, y) = (\frac{f(z)}{2}, y) it follows that
f(z)+f(y)=f(f(z)2/4+f(z)f(y)/2+y)=f(f(y)2/4+f(z)f(y)/2+z) f(z) + f(y) = f(f(z)^2/4 + f(z)f(y)/2 + y) = f(f(y)^2/4 + f(z)f(y)/2 + z)
That is, f(z)2/4−z=Cf(z)^2/4 - z = C. Hence, f(z)=4z+Cf(z) = \sqrt{4z + C}, for some C≥0C \ge 0. □\square

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