Difference between revisions of "2016 APMO Problems/Problem 5"
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==Solution== | ==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. | <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: |
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+ | |||
+ | <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> |
Revision as of 22:43, 12 July 2021
Problem
Find all functions such that
for all positive real numbers
.
Solution
We claim that is the only solution. It is easy to check that it works. Now, we will break things down in several claims. Let
be the assertion to the Functional Equation.
Claim 1: is injective.
Proof: Assume for some
. Now, from
and
we have:
Now comparing, we have as desired.