Difference between revisions of "2024 USAJMO Problems/Problem 5"
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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{ } (3) | + | 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 15:44, 20 October 2024
Contents
[hide]Problem
Find all functions that satisfy for all .
Solution 1
Plugging in as
-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.