Difference between revisions of "2024 USAJMO Problems/Problem 5"

Revision as of 21:15, 23 August 2025

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

Solution 3(Unfinished)

Start with $y = 0$ to get: $f(x^{2}) = f(f(x)) + f(0)$ . Then substitute $x = 0$ to get: $f(-y) + 2yf(0) = f(f(0)) + f(y)$ . We will show that $f(f(0)) = 0$ . Substitute $x = y = 0$ to get $f(0) = f(f(0)) + f(0)$ which yields the desired $f(f(0)) = 0$ . Hence, $f(-x) + 2xf(0) = f(x)$ since $y$ is just a dummy variable. Now we solve for $f(0)$ in our two equations we found. We have that $f(0) = f(x^{2}) - f(f(x))$ from the first equation and that $f(0) = \frac{f(x) - f(-x)}{2x}$ from the second equation.

Hence, we set these equal to get:

$2xf(x^{2}) - 2xf(f(x)) = f(x) - f(-x)$

Replace $x$ with $-x$ to get:

$-2xf(x^{2}) + 2xf(f(-x)) = f(-x) + f(x)$

Now add to get:

$2x(f(f(-x)) - f(f(x))) = 0$

Now since $x$ can't obviously be equal to $0$ , we need $f(f(-x)) = f(f(x))$ .

Case 1: $f$ is an injective function

Hence, $f(-x) = f(x)$ and thus $f(x)$ is an even function.

$\textbf{Claim: f cannot have degree at least 4}$

$\textbf{Proof:}$

Assume $f$ is a polynomial. Assume there exists a solution of which $f$ has degree at least $4$ . Hence, $f(x) = a_n x^{2n} + a_{n - 1} x^{2n - 2} + ... + a_1 x^{2} + a_0$ where $n \geq 2$ . If we go back to the original equation in the problem statement and plug in $y = x^{2}$ , then we have:

$f(0) + 2x^{2} f(x) = f(f(x)) + f(x^{2})$

So we have:

$a_0 + 2x^{2}f(x) = a_n f(x)^{2n} + ... + a_1 f(x)^{2} + a_0$

Hence, $2x^{2} f(x) = a_n f(x)^{2n} + ... + a_1 f(x)^{2}$

But we know $f$ has degree $2n$ so the $LHS$ will have degree $2 + 2n$ but the $RHS$ will have degree $(2n)(2n) = 4n^{2}$ . Solving $4n^{2} = 2n + 2$ will have no integer solutions. Recall that $n \geq 2$ . Clearly, no functions will satisfy this equation and the claim is proven.

We have shown that $f(x)$ can have degree at most $2$ if it is an injective function. Hence, let $f(x) = ax^{2} + bx + c$ . Recall that we had shown earlier that $f(f(0)) = 0$ . Hence, $a f(0)^{2} + b f(0) + c = 0$ but we know $f(0) = c$ . Hence, $ac^{2} + bc + c = 0$ which yields $c(ac + b + 1) = 0$ . Clearly, $c = 0$ has to occur. Hence, $f(x) = ax^{2} + bx$ . Now, recall we showed that $f(x^{2}) = f(f(x)) + f(0)$ . Hence, $ax^{4} + bx^{2} = af(x)^{2} + bf(x) + f(0)$ . Plugging and simplifying, we eventually get:

$x^{4} (a - a^{3}) - 2a^{2} b x^{3} - abx^{2} (b + 1) = f(0) = constant$

Hence, all the $x$ terms must cancel out for that to be a constant. This can be achieved by either $a = 0$ and $b$ can be anything OR $b = 0$ and $a - a^{3} = 0$ . Now if $a = 0$ , then $f(x) = bx$ but recall that $f(x)$ is an even function and hence $b = 0$ is the only suitable option for a linear function to turn into an even function. Hence, $f(x) = 0$ is a solution. If $b = 0$ and $a - a^{3} = 0$ , then $a = 1, -1, 0$ and hence $f(x) = x^{2}, -x^{2}$ are two more new solutions.

Now we have to consider if $f(x)$ is not a polynomial. (not finished yet)

Case 2: $f$ is not an injective function (not finished yet)

~ilikemath247365

@@ Line 126: / Line 126: @@
 Hence, <math>2x^{2} f(x) = a_n f(x)^{2n} + ... + a_1 f(x)^{2}</math>
-But we know <math>f</math> has degree <math>2n</math> so the <math>LHS</math> will have degree <math>2 + 2n</math> but the <math>RHS</math> will have degree <math>(2n)(2n) = 4n^{2}</math>. Solving <math>4n^{2} = 2n + 2 will have no integer solutions. Recall that </math>n \geq 2<math>. Clearly, no functions will satisfy this equation and the claim is proven.
+But we know <math>f</math> has degree <math>2n</math> so the <math>LHS</math> will have degree <math>2 + 2n</math> but the <math>RHS</math> will have degree <math>(2n)(2n) = 4n^{2}</math>. Solving <math>4n^{2} = 2n + 2</math> will have no integer solutions. Recall that <math>n \geq 2</math>. Clearly, no functions will satisfy this equation and the claim is proven.
-We have shown that </math>f(x)<math> can have degree at most </math>2<math> if it is an injective function. Hence, let </math>f(x) = ax^{2} + bx + c<math>. Recall that we had shown earlier that </math>f(f(0)) = 0<math>. Hence, </math>a f(0)^{2} + b f(0) + c = 0<math> but we know </math>f(0) = c<math>. Hence, </math>ac^{2} + bc + c = 0<math> which yields </math>c(ac + b + 1) = 0<math>. Clearly, </math>c = 0<math> has to occur. Hence, </math>f(x) = ax^{2} + bx<math>. Now, recall we showed that </math>f(x^{2}) = f(f(x)) + f(0)<math>. Hence, </math>ax^{4} + bx^{2} = af(x)^{2} + bf(x) + f(0)<math>. Plugging and simplifying, we eventually get:
+We have shown that <math>f(x)</math> can have degree at most <math>2</math> if it is an injective function. Hence, let <math>f(x) = ax^{2} + bx + c</math>. Recall that we had shown earlier that <math>f(f(0)) = 0</math>. Hence, <math>a f(0)^{2} + b f(0) + c = 0</math> but we know <math>f(0) = c</math>. Hence, <math>ac^{2} + bc + c = 0</math> which yields <math>c(ac + b + 1) = 0</math>. Clearly, <math>c = 0</math> has to occur. Hence, <math>f(x) = ax^{2} + bx</math>. Now, recall we showed that <math>f(x^{2}) = f(f(x)) + f(0)</math>. Hence, <math>ax^{4} + bx^{2} = af(x)^{2} + bf(x) + f(0)</math>. Plugging and simplifying, we eventually get:
-</math>x^{4} (a - a^{3}) - 2a^{2} b x^{3} - abx^{2} (b + 1) = f(0) = constant<math>
+<math>x^{4} (a - a^{3}) - 2a^{2} b x^{3} - abx^{2} (b + 1) = f(0) = constant</math>
-Hence, all the </math>x<math> terms must cancel out for that to be a constant. This can be achieved by either </math>a = 0<math> and </math>b<math> can be anything OR </math>b = 0<math> and </math>a - a^{3} = 0<math>. Now if </math>a = 0<math>, then </math>f(x) = bx<math> but recall that </math>f(x)<math> is an even function and hence </math>b = 0<math> is the only suitable option for a linear function to turn into an even function. Hence, </math>f(x) = 0<math> is a solution. If </math>b = 0<math> and </math>a - a^{3} = 0<math>, then </math>a = 1, -1, 0<math> and hence </math>f(x) = x^{2}, -x^{2}<math> are two more new solutions.
+Hence, all the <math>x</math> terms must cancel out for that to be a constant. This can be achieved by either <math>a = 0</math> and <math>b</math> can be anything OR <math>b = 0</math> and <math>a - a^{3} = 0</math>. Now if <math>a = 0</math>, then <math>f(x) = bx</math> but recall that <math>f(x)</math> is an even function and hence <math>b = 0</math> is the only suitable option for a linear function to turn into an even function. Hence, <math>f(x) = 0</math> is a solution. If <math>b = 0</math> and <math>a - a^{3} = 0</math>, then <math>a = 1, -1, 0</math> and hence <math>f(x) = x^{2}, -x^{2}</math> are two more new solutions.
-Now we have to consider if </math>f(x)<math> is not a polynomial. (not finished yet)
+Now we have to consider if <math>f(x)</math> is not a polynomial. (not finished yet)
-Case 2: </math>f$ is not an injective function (not finished yet)
+Case 2: <math>f</math> is not an injective function (not finished yet)

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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