Difference between revisions of "2024 USAJMO Problems/Problem 5"
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== Solution 1 == | == Solution 1 == | ||
+ | Plugging in <math>y</math> as <math>0:</math> | ||
+ | \begin{equation} | ||
+ | f(x^2)=f(f(x))+f(0) \text{ } (1) | ||
+ | \end{equation} | ||
+ | Plugging in <math>x, y</math> as <math>0:</math> | ||
+ | <cmath>f(0)=f(f(0))+f(0)</cmath> | ||
+ | or | ||
+ | <cmath>f(f(0))=0</cmath> | ||
+ | Plugging in <math>x</math> as <math>0:</math> | ||
+ | <cmath>f(-y)+2yf(0)=f(f(0))+f(y),</cmath> | ||
+ | but since <math>f(f(0))=0,</math> | ||
+ | \begin{equation} | ||
+ | f(-y)+2yf(0)=f(y) \text{ } (2) | ||
+ | \end{equation} | ||
+ | Plugging in <math>y^2</math> instead of <math>y</math> in the given equation: | ||
+ | <cmath>f(x^2-y^2)+2y^2f(x)=f(f(x))+f(y^2)</cmath> | ||
+ | Replacing <math>y</math> and <math>x</math>: | ||
+ | <cmath>f(y^2-x^2)+2x^2f(y)=f(f(y))+f(x^2)</cmath> | ||
+ | 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 <math>f(0)-f(0)=0</math> by <math>(1).</math> Also, | ||
+ | <cmath>f(x^2-y^2)-f(y^2-x^2)=2(x^2-y^2)f(0)</cmath> by <math>(2)</math> | ||
+ | So, <math>(3)</math> is reduced to: | ||
+ | <cmath>2(x^2-y^2)f(0)+2y^2f(x)-2x^2f(y)=0</cmath> | ||
+ | Regrouping and dividing by 2: | ||
+ | <cmath>y^2(f(x)-f(0))=x^2(f(y)-f(0))</cmath> | ||
+ | <cmath>\frac{f(x)-f(0)}{x^2}=\frac{f(y)-f(0)}{y^2}</cmath> | ||
+ | Because this holds for all x and y, <math>\frac{f(x)-f(0)}{x^2}</math> is a constant. So, <math>f(x)=cx^2+f(0)</math>. | ||
+ | This function must be even, so <math>f(y)-f(-y)=0</math>. | ||
+ | So, along with <math>(2)</math>, <math>2yf(0)=0</math> for all <math>y</math>, so <math>f(0)=0</math>, and <math>f(x)=cx^2</math>. | ||
+ | Plugging in <math>cx^2</math> for <math>f(x)</math> in the original equation, we get: | ||
+ | <cmath>c(x^4-2x^2y+y^2)+2cx^2y=c^3x^4+cy^2</cmath> | ||
+ | <cmath>c(x^4+y^2)=c(c^2x^4+y^2)</cmath> | ||
+ | So, <math>c=0</math> or <math>c^2=1.</math> | ||
+ | All of these solutions work, so the solutions are <math>f(x)=-x^2, 0, x^2</math>. | ||
− | + | -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( | + | <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>. | ||
− | <cmath>f( | + | 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>. | |
− | <cmath>f( | + | 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 | + | <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> | + | <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 16:57, 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.