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, | ||
Line 46: | Line 46: | ||
-codemaster11 | -codemaster11 | ||
+ | |||
+ | == Solution 2 == | ||
+ | |||
+ | Let our equation be <math>P(x,y)</math>. We start by plugging in some initial values: | ||
+ | |||
+ | <math>y=x^2:\; f(0)+2x^2f(x) = f(f(x))+f(x^2) \;\;\;\; (1)</math> | ||
+ | |||
+ | <math>y=0:\; f(x^2) = f(f(x))+f(0) \;\;\;\; (2)</math> | ||
+ | |||
+ | <math>x=0:\; f(-y) + 2yf(0) = f(f(0)) + f(y) \;\;\;\; (3)</math> | ||
+ | |||
+ | Plugging in <math>x=1</math> into <math>(3)</math> gives | ||
+ | <cmath>f(1) = f(f(1)) + f(0) \;\;\;\; (4).</cmath> | ||
+ | From <math>(1)</math>, we get | ||
+ | <cmath>f(0) + 2x^2f(x) = 2f(x^2)-f(0) \implies x^2f(x)+f(0) = f(x^2)</cmath> | ||
+ | Substituting in what we have in <math>(3)</math> gives | ||
+ | <cmath>x^2f(x)+f(0) = f(0) = f(f(x)) \implies x^2f(x) = f(f(x)).</cmath> | ||
+ | Plugging in <math>x=1</math> gives | ||
+ | <cmath>f(1)=f(f(1)) \;\;\;\; (5).</cmath> | ||
+ | As a result, <math>(4)</math> becomes <math>f(0)=0</math>. | ||
+ | |||
+ | Now, <math>(3)</math> becomes | ||
+ | <cmath>f(x^2) = f(f(x)) \;\;\;\; (6)</cmath> | ||
+ | and <math>(2)</math> becomes | ||
+ | <cmath>f(y)=f(-y) \;\;\;\; (7).</cmath> | ||
+ | Note that <math>f\equiv 0</math> is a solution. Now, assume <math>f(x) \neq 0</math>. | ||
+ | |||
+ | <b>Claim:</b> <math>f</math> is injective over <math>\mathbb{R}^{+}</math>. | ||
+ | |||
+ | Let <math>f(a) = f(b) \neq 0</math> with <math>a,b>0</math>. Plugging in <math>x=a, y=b^2</math> and <math>x=b, y=a^2</math> into <math>P</math> gives us | ||
+ | <cmath>f(a^2-b^2)+2b^2f(a) = f(a^2)+f(b^2)</cmath> | ||
+ | <cmath>f(b^2-a^2)+2a^2f(b) = f(b^2)+f(a^2)</cmath> | ||
+ | Subtracting, and using <math>(7)</math> gives us <math>2(a^2-b^2)f(a) = 0</math>, which implies that either <math>f(a)=0</math> or <math>a=\pm b</math>. Either way leads to contradiction. Thus, <math>f</math> is injective. <math>\square</math> | ||
+ | |||
+ | As a result, <math>(6)</math> becomes <math>f(x)=\pm x^2</math>. Piecing everything yields <math>f(x) = 0, \pm x^2</math>. | ||
+ | |||
+ | It just remains to verify these solutions work, and doing so is quite trivial; | ||
+ | <cmath>f(x)=0:\; 0+0 = 0+0,</cmath> | ||
+ | <cmath>f(x)=x^2:\; (x^2-y)^2+2yx^2 = x^4+y^2,</cmath> | ||
+ | <cmath>f(x)=-x^2:\; -(x^2-y)^2-2yx^2 = -x^4-y^2,</cmath> | ||
+ | all of which are obviously true. | ||
+ | |||
+ | ~sml1809 | ||
==See Also== | ==See Also== | ||
{{USAJMO newbox|year=2024|num-b=4|num-a=6}} | {{USAJMO newbox|year=2024|num-b=4|num-a=6}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Latest revision as of 16:44, 20 October 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.