2024 USAJMO Problems/Problem 5
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 |
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