Difference between revisions of "2014 USAMO Problems/Problem 2"
(→Alternative Solution) |
|||
Line 39: | Line 39: | ||
==Alternative Solution== | ==Alternative Solution== | ||
− | Given that the range of f consists entirely of integers, it is clear that the LHS must be an integer and <math>f(yf(y))</math> must also be an integer, therefore <math>\frac{f(x)^2}{x}</math> is an integer. If <math>x</math> divides <math>f(x)^2</math> for all integers <math>x \ne 0</math>, then <math>x</math> must be a factor of <math>f(x)</math>, therefore <math>f(0)=0</math>. Now, by setting <math>y=0</math> in the original equation, | + | Given that the range of f consists entirely of integers, it is clear that the LHS must be an integer and <math>f(yf(y))</math> must also be an integer, therefore <math>\frac{f(x)^2}{x}</math> is an integer. If <math>x</math> divides <math>f(x)^2</math> for all integers <math>x \ne 0</math>, then <math>x</math> must be a factor of <math>f(x)</math>, therefore <math>f(0)=0</math>. Now, by setting <math>y=0</math> in the original equation, this simplifies to <math>xf(-x)=\frac{f(x)^2}{x}</math>. Assuming <math>x \ne 0</math>, we have <math>x^2f(-x)=f(x)^2</math>. Substituting in <math>-x</math> for <math>x</math> gives us <math>x^2f(x)=f(-x)^2</math>. Substituting in <math>\frac{f(x)^2}{x^2}</math> in for <math>f(-x)</math> in the second equation gives us <math>x^2f(x)=\frac{f(x)^4}{x^4}</math>, so <math>x^6f(x)=f(x)^4</math>. In particular, if <math>f(x) \ne 0</math>, then we have <math>f(x)^3=x^6</math>, therefore <math>f(x)=0, x^2</math> for every <math>x</math>. Now, we just have to prove that if for some integer <math>t \ne 0</math>, if <math>f(t)=0</math>, then <math>f(x)=0</math> for all integers <math>x</math>. If we assume <math>f(y)=0</math> and <math>y \ne 0</math> in the original equation, this simplifies to <math>xf(-x)+y^2f(2x)=\frac{f(x)^2}{x}. However, since </math>x^2f(-x)=f(x)^2<math>, we can rewrite this equation as </math>\frac{f(x)^2}{x}+y^2f(2x)=\frac{f(x)^2}{x}<math>, </math>y^2f(2x)<math> must therefore be equivalent to </math>0<math>. Since, by our initial assumption, </math>y \ne 0<math>, this means that </math>f(2x)=0<math>, so, if for some integer </math>y \ne 0<math>, </math>f(y)=0<math>, then </math>f(x)=0<math> for all integers </math>x<math>. The contrapositive must also be true, i.e. If </math>f(x) \ne 0<math> for all integers </math>x<math>, then there is no integral value of </math>y \ne 0<math> such that </math>f(y)=0<math>, therefore </math>f(x)<math> must be equivalent for </math>x^2<math> for every integer </math>x<math>, including </math>0<math>, since </math>f(0)=0<math>. Thus, </math>f(x)=0, x^2$ are the only possible solutions. |
[[Category:Olympiad Algebra Problems]] | [[Category:Olympiad Algebra Problems]] | ||
[[Category:Functional Equation Problems]] | [[Category:Functional Equation Problems]] |
Revision as of 11:04, 4 January 2019
Problem
Let be the set of integers. Find all functions
such that
for all
with
.
Solution
Note: This solution is kind of rough. I didn't want to put my 7-page solution all over again. It would be nice if someone could edit in the details of the expansions.
Lemma 1: .
Proof: Assume the opposite for a contradiction. Plug in
(because we assumed that
),
. What you get eventually reduces to:
which is a contradiction since the LHS is divisible by 2 but not 4.
Then plug in into the original equation and simplify by Lemma 1. We get:
Then:
Therefore, must be 0 or
.
Now either is
for all
or there exists
such that
. The first case gives a valid solution. In the second case, we let
in the original equation and simplify to get:
But we know that
, so:
Since
is not 0,
is 0 for all
(including 0). Now either
is 0 for all
, or there exists some
such that
. Then
must be odd. We can let
in the original equation, and since
is 0 for all
, stuff cancels and we get:
[b]for
.[/b]
Now, let
and we get:
Now, either both sides are 0 or both are equal to
. If both are
then:
which simplifies to:
Since
and
is odd, both cases are impossible, so we must have:
Then we can let
be anything except 0, and get
is 0 for all
except
. Also since
, we have
, so
is 0 for all
except
. So
is 0 for all
except
. Since
,
. Squaring,
and dividing by
,
. Since
,
, which is a contradiction for
. However, if we plug in
with
and
as an arbitrary large number with
into the original equation, we get
which is a clear contradiction, so our only solutions are
and
.
Alternative Solution
Given that the range of f consists entirely of integers, it is clear that the LHS must be an integer and must also be an integer, therefore
is an integer. If
divides
for all integers
, then
must be a factor of
, therefore
. Now, by setting
in the original equation, this simplifies to
. Assuming
, we have
. Substituting in
for
gives us
. Substituting in
in for
in the second equation gives us
, so
. In particular, if
, then we have
, therefore
for every
. Now, we just have to prove that if for some integer
, if
, then
for all integers
. If we assume
and
in the original equation, this simplifies to
x^2f(-x)=f(x)^2
\frac{f(x)^2}{x}+y^2f(2x)=\frac{f(x)^2}{x}
y^2f(2x)
0
y \ne 0
f(2x)=0
y \ne 0
f(y)=0
f(x)=0
x
f(x) \ne 0
x
y \ne 0
f(y)=0
f(x)
x^2
x
0
f(0)=0
f(x)=0, x^2$ are the only possible solutions.