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) | + | 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, |
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.