Difference between revisions of "2016 APMO Problems/Problem 5"

Revision as of 23:43, 12 July 2021

Problem

Find all functions $f: \mathbb{R}^+ \to \mathbb{R}^+$ such that $(z + 1)f(x + y) = f(xf(z) + y) + f(yf(z) + x),$ for all positive real numbers $x, y, z$ .

Solution

We claim that $f(x)=x$ is the only solution. It is easy to check that it works. Now, we will break things down in several claims. Let $P(x.y,z)$ be the assertion to the Functional Equation.

Claim 1: $f$ is injective.

Proof: Assume $f(a)=f(b)$ for some $a,b \in \mathbb{R}^+$ . Now, from $P(x,y,a)$ and $P(x,y,b$ we have:

$(a+1)f(x+y)=f(xf(a)+y)+f(yf(a)+x)$ $(b+1)f(x+y)=f(xf(b)+y)+f(yf(b)+x)$

Now comparing, we have $a=b$ as desired. $\square$

@@ Line 5: / Line 5: @@
 ==Solution==
-We claim that <math>f(x)=x</math> is the only solution. It is easy to check that it works. Now, we will break things down in several claims.
+We claim that <math>f(x)=x</math> is the only solution. It is easy to check that it works. Now, we will break things down in several claims. Let <math>P(x.y,z)</math> be the assertion to the Functional Equation.
 <b>Claim 1:</b> <math>f</math> is injective.
-<i>Proof:</i>
+<i>Proof:</i> Assume <math>f(a)=f(b)</math> for some <math>a,b \in \mathbb{R}^+</math>. Now, from <math>P(x,y,a)</math> and <math>P(x,y,b</math> we have:
+<cmath>(a+1)f(x+y)=f(xf(a)+y)+f(yf(a)+x)</cmath>
+<cmath>(b+1)f(x+y)=f(xf(b)+y)+f(yf(b)+x)</cmath>
+Now comparing, we have <math>a=b</math> as desired. <math>\square</math>

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Difference between revisions of "2016 APMO Problems/Problem 5"

Revision as of 23:43, 12 July 2021

Problem

Solution