Difference between revisions of "2016 APMO Problems/Problem 5"
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− | This gives us the power to compute <math>f(1)</math>. From <math>P(1,1,1)</math> we get <math>f(f(1)+1)=f(2)</math> and injectivity gives <math>f(1)=1</math>. | + | This gives us the power to compute <math>f(1)</math>. From <math>P(1,1,1)</math> we get <math>f(f(1)+1)=f(2)</math> and injectivity gives <math>f(1)=1</math>. Showing that <math>f</math> is unbounded above is also easy as we can fix <math>(x,y)</math> and let <math>z</math> blow up to <math>\infty</math>. |
Revision as of 05:43, 13 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.
This gives us the power to compute . From
we get
and injectivity gives
. Showing that
is unbounded above is also easy as we can fix
and let
blow up to
.
Claim 2: is surjective.
Proof: gives
This gives that . Putting
, we get
. By induction, surjectivity is proved as
and we are essentially done.