2024 USAJMO Problems/Problem 5

Problem

Find all functions $f:\mathbb{R}\rightarrow\mathbb{R}$ that satisfy $f(x^2-y)+2yf(x)=f(f(x))+f(y)$ for all $x,y\in\mathbb{R}$ .

Solution 1

Plugging in $y$ as $0:$ \begin{equation} f(x^2)=f(f(x))+f(0) \text{ } (1) \end{equation} Plugging in $x, y$ as $0:$ $f(0)=f(f(0))+f(0)$ or $f(f(0))=0$ Plugging in $x$ as $0:$ $f(-y)+2yf(0)=f(f(0))+f(y),$ but since $f(f(0))=0,$ \begin{equation} f(-y)+2yf(0)=f(y) \text{ } (2) \end{equation} Plugging in $y^2$ instead of $y$ in the given equation: $f(x^2-y^2)+2y^2f(x)=f(f(x))+f(y^2)$ Replacing $y$ and $x$ : $f(y^2-x^2)+2x^2f(y)=f(f(y))+f(x^2)$ 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 $f(0)-f(0)=0$ by $(1).$ Also, $f(x^2-y^2)-f(y^2-x^2)=2(x^2-y^2)f(0)$ by $(2)$ So, $(3)$ is reduced to: $2(x^2-y^2)f(0)+2y^2f(x)-2x^2f(y)=0$ Regrouping and dividing by 2: $y^2(f(x)-f(0))=x^2(f(y)-f(0))$ $\frac{f(x)-f(0)}{x^2}=\frac{f(y)-f(0)}{y^2}$ Because this holds for all x and y, $\frac{f(x)-f(0)}{x^2}$ is a constant. So, $f(x)=cx^2+f(0)$ . This function must be even, so $f(y)-f(-y)=0$ . So, along with $(2)$ , $2yf(0)=0$ for all $y$ , so $f(0)=0$ , and $f(x)=cx^2$ . Plugging in $cx^2$ for $f(x)$ in the original equation, we get: $c(x^4-2x^2y+y^2)+2cx^2y=c^3x^4+cy^2$ $c(x^4+y^2)=c(c^2x^4+y^2)$ So, $c=0$ or $c^2=1.$ All of these solutions work, so the solutions are $f(x)=-x^2, 0, x^2$ .

-codemaster11

Solution 2

Let our equation be $P(x,y)$ . We start by plugging in some initial values:

$y=x^2:\; f(0)+2x^2f(x) = f(f(x))+f(x^2) \;\;\;\; (1)$

$y=0:\; f(x^2) = f(f(x))+f(0) \;\;\;\; (2)$

$x=0:\; f(-y) + 2yf(0) = f(f(0)) + f(y) \;\;\;\; (3)$

Plugging in $x=1$ into $(3)$ gives $f(1) = f(f(1)) + f(0) \;\;\;\; (4).$ From $(1)$ , we get $f(0) + 2x^2f(x) = 2f(x^2)-f(0) \implies x^2f(x)+f(0) = f(x^2)$ Substituting in what we have in $(3)$ gives $x^2f(x)+f(0) = f(0) = f(f(x)) \implies x^2f(x) = f(f(x)).$ Plugging in $x=1$ gives $f(1)=f(f(1)) \;\;\;\; (5).$ As a result, $(4)$ becomes $f(0)=0$ .

Now, $(3)$ becomes $f(x^2) = f(f(x)) \;\;\;\; (6)$ and $(2)$ becomes $f(y)=f(-y) \;\;\;\; (7).$ Note that $f\equiv 0$ is a solution. Now, assume $f(x) \neq 0$ .

Claim: $f$ is injective over $\mathbb{R}^{+}$ .

Let $f(a) = f(b) \neq 0$ with $a,b>0$ . Plugging in $x=a, y=b^2$ and $x=b, y=a^2$ into $P$ gives us $f(a^2-b^2)+2b^2f(a) = f(a^2)+f(b^2)$ $f(b^2-a^2)+2a^2f(b) = f(b^2)+f(a^2)$ Subtracting, and using $(7)$ gives us $2(a^2-b^2)f(a) = 0$ , which implies that either $f(a)=0$ or $a=\pm b$ . Either way leads to contradiction. Thus, $f$ is injective. $\square$

As a result, $(6)$ becomes $f(x)=\pm x^2$ . Piecing everything yields $f(x) = 0, \pm x^2$ .

It just remains to verify these solutions work, and doing so is quite trivial; $f(x)=0:\; 0+0 = 0+0,$ $f(x)=x^2:\; (x^2-y)^2+2yx^2 = x^4+y^2,$ $f(x)=-x^2:\; -(x^2-y)^2-2yx^2 = -x^4-y^2,$ all of which are obviously true.

~sml1809

2024 USAJMO (Problems • Resources)
Preceded by Problem 4	Followed by Problem 6
1 • 2 • 3 • 4 • 5 • 6
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2024 USAJMO Problems/Problem 5

Contents

Problem

Solution 1

Solution 2

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