Difference between revisions of "2024 USAJMO Problems/Problem 5"
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Plugging in <math>y</math> as <math>0:</math> | Plugging in <math>y</math> as <math>0:</math> | ||
\begin{equation} | \begin{equation} | ||
− | f(x^2)=f(f(x))+f(0) \text{ } | + | f(x^2)=f(f(x))+f(0) \text{ } (1) |
\end{equation} | \end{equation} | ||
Plugging in <math>x, y</math> as <math>0:</math> | Plugging in <math>x, y</math> as <math>0:</math> | ||
Line 19: | Line 19: | ||
but since <math>f(f(0))=0,</math> | but since <math>f(f(0))=0,</math> | ||
\begin{equation} | \begin{equation} | ||
− | f(-y)+2yf(0)=f(y) \text{ } | + | f(-y)+2yf(0)=f(y) \text{ } (2) |
\end{equation} | \end{equation} | ||
Plugging in <math>y^2</math> instead of <math>y</math> in the given equation: | Plugging in <math>y^2</math> instead of <math>y</math> in the given equation: | ||
Line 27: | Line 27: | ||
The difference would be: | The difference would be: | ||
\begin{equation} | \begin{equation} | ||
− | f(x^2-y^2)-f(y^2-x^2)+2y^2f(x)-2x^2f(y)=f(f(x))-f(x^2)-f(f(y))-f(y^2) \text{ } | + | f(x^2-y^2)-f(y^2-x^2)+2y^2f(x)-2x^2f(y)=f(f(x))-f(x^2)-f(f(y))-f(y^2) \text{ } (3) |
\end{equation} | \end{equation} | ||
The right-hand side would be <math>f(0)-f(0)=0</math> by <math>(1).</math> Also, | The right-hand side would be <math>f(0)-f(0)=0</math> by <math>(1).</math> Also, |
Latest revision as of 16:57, 20 August 2024
Contents
Problem
Find all functions that satisfy
for all
.
Solution 1
Plugging in as
\begin{equation}
f(x^2)=f(f(x))+f(0) \text{ } (1)
\end{equation}
Plugging in
as
or
Plugging in
as
but since
\begin{equation}
f(-y)+2yf(0)=f(y) \text{ } (2)
\end{equation}
Plugging in
instead of
in the given equation:
Replacing
and
:
The difference would be:
\begin{equation}
f(x^2-y^2)-f(y^2-x^2)+2y^2f(x)-2x^2f(y)=f(f(x))-f(x^2)-f(f(y))-f(y^2) \text{ } (3)
\end{equation}
The right-hand side would be
by
Also,
by
So,
is reduced to:
Regrouping and dividing by 2:
Because this holds for all x and y,
is a constant. So,
.
This function must be even, so
.
So, along with
,
for all
, so
, and
.
Plugging in
for
in the original equation, we get:
So,
or
All of these solutions work, so the solutions are
.
-codemaster11
Solution 2
Let our equation be . We start by plugging in some initial values:
Plugging in into
gives
From
, we get
Substituting in what we have in
gives
Plugging in
gives
As a result,
becomes
.
Now, becomes
and
becomes
Note that
is a solution. Now, assume
.
Claim: is injective over
.
Let with
. Plugging in
and
into
gives us
Subtracting, and using
gives us
, which implies that either
or
. Either way leads to contradiction. Thus,
is injective.
As a result, becomes
. Piecing everything yields
.
It just remains to verify these solutions work, and doing so is quite trivial;
all of which are obviously true.
~sml1809
See Also
2024 USAJMO (Problems • Resources) | ||
Preceded by Problem 4 |
Followed by Problem 6 | |
1 • 2 • 3 • 4 • 5 • 6 | ||
All USAJMO Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.