USAMO

Algebra Difficulty 7.5 National Olympiad, round 2 Find the answer

Let $\mathbb{R}^{+}$ be the set of positive real numbers. Find all functions $f:\mathbb{R}^{+}\rightarrow\mathbb{R}^{+}$ such that, for all $x, y \in \mathbb{R}^{+}$ , $f(xy + f(x)) = xf(y) + 2$

Your answer

A number or a short expression. Fractions can be typed as 3/2, and spacing doesn't matter.

Solution

Make the following substitutions to the equation:
1. $(x, 1) \rightarrow f(x + f(x)) = xf(1) + 2$
2. $(1, x + f(x)) \rightarrow f(x + f(x) + f(1)) = f(x + f(x)) + 2 = xf(1) + 4$
3. $(x, 1 + \frac{f(1)}{x}) \rightarrow f(x + f(x) + f(1)) = xf\biggl(1 + \frac{f(1)}{x}\biggr) + 2$
It then follows from (2) and (3) that $f(1 + \frac{f(1)}{x}) = f(1) + \frac{2}{x}$ , so we know that this function is linear for $x > 1$ . Substitute $f(x) = ax+b$ and solve for $a$ and $b$ in the functional equation; we find that $f(x) = x + 1 \forall x > 1$ .
Now, we can let $x > 1$ and $y \le 1$ . Since $f(x) = x + 1$ , $xy + f(x) > x > 1$ , so $f(xy + f(x)) = xy + x + 2 = xf(y) + 2$ . It becomes clear then that $f(y) = y + 1$ as well, so $f(x) = x + 1$ is the only solution to the functional equation.
~jkmmm3

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