2024 USAJMO Problems/Problem 5: Difference between revisions

Latest revision as of 22:27, 11 November 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

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.

Claim: $f$ cannot have degree at least $4$

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. $\square$

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.

Claim: $f$ has to be a polynomial to satisfy the given problem statement

Proof:

We will first show that $f(f(x)) = x^{2} f(x)$ . Substitute $y = \frac{x^{2}}{2}$ :

$f(\frac{x^{2}}{2}) + x^{2} f(x) = f(f(x)) + f(\frac{x^{2}}{2})$ . Hence, the lemma. $\square$

We have already shown that if $f$ is injective, then $f$ is even and this proof lies in the case of when $f$ is injective. Hence, $f(x)$ is an even function. Recall that $f(x^{2}) = f(f(x)) + f(0)$ . Hence, $f(x^{2}) = x^{2} f(x) + f(0)$ . Now we plug in $x = 1$ to get $f(1) = f(1) + f(0)$ and hence $f(0) = 0$ . In other words, $f(x^{2}) = x^{2} f(x)$ . (Haven't finished proof yet)

Case 2: $f$ is not an injective function

We can prove that $f$ has to be injective. This is shown in sml1809's solution above in which we claim $f(a) = f(b) \neq 0$ and substitute $x = a, y = b^{2}$ and $x = b, y = a^{2}$ to prove the claim that $f$ has to be injective.

~ilikemath247365

Solution 4(Another solution)

We will show $x^2 f(x) = f(f(x))$ . First, the original equation is

$f(x^2 - y) + 2yf(x) = f(f(x)) + f(y)$

We will make the substitution $y = x^2 - 1$

$f(1) + 2(x^2 - 1)f(x) = f(f(x)) + f(x^2 - 1)$

We will now take the original equation again and substitute $y = 1$

$f(x^2 - 1) + 2f(x) = f(f(x)) + f(1) \implies f(x^2 - 1) = f(f(x)) + f(1) - 2f(x)$

Now we will substitute this new representation of $f(x^2 - 1)$ into our equation to get

$f(1) + (2x^2 - 2)f(x) = f(f(x) + f(f(x)) + f(1) - 2f(x) \implies (2x^2)f(x) = 2f(f(x)) \implies x^2 f(x) = f(f(x)) \square.$

We will now plug this in to our given equation

$f(x^2 - y) + 2yf(x) = f(f(x)) + f(y) \implies f(x^2 - y) + 2yf(x) = x^2 f(x) + f(y) \implies f(x^2 - y) - f(y) = f(x)(x^2 - 2y)$ .

Now we will make the substitution $x = y$

$f(x^2 - x) - f(x) = f(x)(x^2 - 2x) \implies f(x^2 - x) = f(x)(x^2 - 2x + 1) \implies f(x(x - 1)) = f(x)(x - 1)^2$ . This immediately raises a conjecture. If we have two real numbers $a, b$ where $f$ is defined at these two real numbers, then

$f(ab) = f(a) b^2$

Let's prove this. From our equation $f(x^2 - y) + 2yf(x) = x^2 f(x) + f(y)$ , we can substitute $y = -x$ to get

$f(x^2 + x) - 2xf(x) = x^2 f(x) + f(-x) \implies f(x^2 + x) = f(x)(x^2 + 2x) + f(-x)$ . If we can show $f(-x) = f(x)$ , then $f(x^2 + x) = f(x(x + 1)) = f(x)(x^2 + 2x + 1) = f(x) (x + 1)^2 \implies f(x(x + 1) = f(x) (x + 1)^2$ and we should be done as we can now prove $f(ab) = f(a) b^2$ by induction. So let's prove $f(-x) = f(x)$ .

The proof is simple. From our given equation $f(x^2 - y) + 2yf(x) = f(f(x)) + f(y)$ , let's substitute $x = 0$ to get

$f(-y) = f(f(0)) + f(y)$ .

We will prove $f(0) = 0$ . From our given equation $f(x^2 - y) + 2yf(x) = f(f(x)) + f(y)$ , we will plug in $x = y = 1$ to get

$f(0) + 2f(1) = f(f(1)) + f(1) \implies f(0) + f(1) = f(f(1)) \implies f(0) + f(1) = 1^2 f(1) \implies f(0) = 0$ following $f(f(x)) = x^2 f(x)$ . Henceforth, we go back to $f(-y) = f(f(0)) + f(y) \implies f(-y) = f(0) + f(y) \implies f(-y) = f(y) \implies f(-x) = f(x) \square.$

Hence, as $f(-x) = f(x)$ , we had concluded before that if this was the case,

$f(x(x + 1)) = f(x) (x + 1)^2$ and $f(x(x - 1)) = f(x) (x - 1)^2$ where the second equation we had derived before proving $f(-x) = f(x)$ . From these two equations, we can generate any two real numbers that differ by $1$ where

$f(ab) = f(a) b^2$ where $a - b = 1$ .

We will show this happens for all real $a, b$ as we have proved only part of our conjecture. Note that $f(a) = f(-a)$ has been proved.

We will assume $f(ab) = f(a) b^2$ only holds true if and only if $a - b = 1$ . Now let $b = -1$ . We have

$f(-a) = f(a)$ which holds true for all real numbers. However, we assumed the above was true if and only if $a - b = 1$ . This would imply the above was true if and only if $a = 0$ which is not the case. Hence, this is a contradiction and we have proved our conjecture. $\square.$

Now, we have shown $f(ab) = f(a) b^2$ is true for all real $a, b$ . Let $a = 1$ to get

$f(b) = b^2$ which means $f(x) = x^2$ is a solution.

Now consider $ab = x \implies b = \frac{x}{a}$ . Now $f(x) = f(a) \frac{x^2}{a^2} \implies \frac{f(x)}{x^2} = \frac{f(a)}{a^2}$ . From here we can generate many quadratic functions. Knowing $f(x) = x^2$ is a solution gives us motivation for $f(x) = -x^2$ which we can confirm will satisfy the given equation. We can prove this also

Given the original equation $f(x^2 - y) + 2yf(x) = f(f(x)) + f(y)$ , we have found $f(x) = x^2$ is a solution. We will show that if $f(x)$ is a solution, so is $-f(x)$ . First, we'll assume something different, then prove by contradiction. We claim that if $f(x)$ is a solution, $-f(x)$ is not a solution. Let's plug in $f(x) = x^2$ since we already know that's a solution. We have

$(x^2 - y)^2 + 2x^2 y = x^4 + y^2 \implies x^4 - 2x^2 y + y^2 + 2x^2 y = x^4 + y^2$ which holds.

So now since we know $f(x) = x^2$ is a solution, we had claimed that $-f(x) = -x^2$ was not a solution. Hence we can plug this in to get

$-(x^2 - y)^2 - 2x^2 y = f(-x^2) - y^2 \implies -(x^2 - y)^2 - 2x^2 y = -x^4 - y^2$ of which this is exactly the negation of the equation we got for $f(x) = x^2$ . Therefore $-f(x)$ is indeed a solution and we have a contradiction with $-f(x)$ not being a solution. Thus $-f(x)$ has to be a solution. $\square$ .

Now assuming $f$ is constant, we have

$f(x) = c \implies c + 2yc = x^2 c + c \implies 2yc = x^2 c \implies c = 0$ is a solution. Hence $f(x) = 0$ is also a solution and so $f(x) = x^2, -x^2, 0$

~ilikemath247365

@@ Line 90: / Line 90: @@
 ~sml1809
-==Solution 3(Unfinished)==
+==Solution 3==
-Start with <math>y = 0</math> to get: <math>f(x^{2}) = f(f(x)) + f(0)</math>. Then substitute <math>x = 0</math> to get: <math>f(-y) + 2yf(0) = f(f(0)) + f(y)</math>. We will show that <math>f(f(0)) = 0</math>. Substitute <math>x = y = 0</math> to get <math>f(0) = f(f(0)) + f(0)</math> which yields the desired <math>f(f(0)) = 0</math>. Hence, <math>f(-x) + 2xf(0) = f(x)</math> since <math>y</math> is just a dummy variable. Now we solve for <math>f(0)</math> in our two equations we found. We have that <math>f(0) = f(x^{2}) - f(f(x))</math> from the first equation and that <math>f(0) = \frac{f(x) - f(-x)}{2x}</math> from the second equation.
+Start with <imath>y = 0</imath> to get: <imath>f(x^{2}) = f(f(x)) + f(0)</imath>. Then substitute <imath>x = 0</imath> to get: <imath>f(-y) + 2yf(0) = f(f(0)) + f(y)</imath>. We will show that <imath>f(f(0)) = 0</imath>. Substitute <imath>x = y = 0</imath> to get <imath>f(0) = f(f(0)) + f(0)</imath> which yields the desired <imath>f(f(0)) = 0</imath>. Hence, <imath>f(-x) + 2xf(0) = f(x)</imath> since <imath>y</imath> is just a dummy variable. Now we solve for <imath>f(0)</imath> in our two equations we found. We have that <imath>f(0) = f(x^{2}) - f(f(x))</imath> from the first equation and that <imath>f(0) = \frac{f(x) - f(-x)}{2x}</imath> from the second equation.
 Hence, we set these equal to get:
-<math>2xf(x^{2}) - 2xf(f(x)) = f(x) - f(-x)</math>
+<imath>2xf(x^{2}) - 2xf(f(x)) = f(x) - f(-x)</imath>
-Replace <math>x</math> with <math>-x</math> to get:
+Replace <imath>x</imath> with <imath>-x</imath> to get:
-<math>-2xf(x^{2}) + 2xf(f(-x)) = f(-x) + f(x)</math>
+<imath>-2xf(x^{2}) + 2xf(f(-x)) = f(-x) + f(x)</imath>
 Now add to get:
-<math>2x(f(f(-x)) - f(f(x))) = 0</math>
+<imath>2x(f(f(-x)) - f(f(x))) = 0</imath>
-Now since <math>x</math> can't obviously be equal to <math>0</math>, we need <math>f(f(-x)) = f(f(x))</math>.
+Now since <imath>x</imath> can't obviously be equal to <imath>0</imath>, we need <imath>f(f(-x)) = f(f(x))</imath>.
-Case 1: <math>f</math> is an injective function
+Case 1: <imath>f</imath> is an injective function
-Hence, <math>f(-x) = f(x)</math> and thus <math>f(x)</math> is an even function.
+Hence, <imath>f(-x) = f(x)</imath> and thus <imath>f(x)</imath> is an even function.
-<math>\textbf{Claim: f cannot have degree at least 4}</math>
+<b>Claim:</b> <imath>f</imath> cannot have degree at least <imath>4</imath>
-<math>\textbf{Proof:}</math>
+<b>Proof:</b>
-Assume <math>f</math> is a polynomial. Assume there exists a solution of which <math>f</math> has degree at least <math>4</math>. Hence, <math>f(x) = a_n x^{2n} + a_{n - 1} x^{2n - 2} + ... + a_1 x^{2} + a_0</math> where <math>n \geq 2</math>. If we go back to the original equation in the problem statement and plug in <math>y = x^{2}</math>, then we have:
+Assume <imath>f</imath> is a polynomial. Assume there exists a solution of which <imath>f</imath> has degree at least <imath>4</imath>. Hence, <imath>f(x) = a_n x^{2n} + a_{n - 1} x^{2n - 2} + ... + a_1 x^{2} + a_0</imath> where <imath>n \geq 2</imath>. If we go back to the original equation in the problem statement and plug in <imath>y = x^{2}</imath>, then we have:
-<math>f(0) + 2x^{2} f(x) = f(f(x)) + f(x^{2})</math>
+<imath>f(0) + 2x^{2} f(x) = f(f(x)) + f(x^{2})</imath>
 So we have:
-<math>a_0 + 2x^{2}f(x) = a_n f(x)^{2n} + ... + a_1 f(x)^{2} + a_0</math>
+<imath>a_0 + 2x^{2}f(x) = a_n f(x)^{2n} + ... + a_1 f(x)^{2} + a_0</imath>
-Hence, <math>2x^{2} f(x) = a_n f(x)^{2n} + ... + a_1 f(x)^{2}</math>
+Hence, <imath>2x^{2} f(x) = a_n f(x)^{2n} + ... + a_1 f(x)^{2}</imath>
-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. <math>\square</math>
+But we know <imath>f</imath> has degree <imath>2n</imath> so the <imath>LHS</imath> will have degree <imath>2 + 2n</imath> but the <imath>RHS</imath> will have degree <imath>(2n)(2n) = 4n^{2}</imath>. Solving <imath>4n^{2} = 2n + 2</imath> will have no integer solutions. Recall that <imath>n \geq 2</imath>. Clearly, no functions will satisfy this equation and the claim is proven. <imath>\square</imath>
-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 <imath>f(x)</imath> can have degree at most <imath>2</imath> if it is an injective function. Hence, let <imath>f(x) = ax^{2} + bx + c</imath>. Recall that we had shown earlier that <imath>f(f(0)) = 0</imath>. Hence, <imath>a f(0)^{2} + b f(0) + c = 0</imath> but we know <imath>f(0) = c</imath>. Hence, <imath>ac^{2} + bc + c = 0</imath> which yields <imath>c(ac + b + 1) = 0</imath>. Clearly, <imath>c = 0</imath> has to occur. Hence, <imath>f(x) = ax^{2} + bx</imath>. Now, recall we showed that <imath>f(x^{2}) = f(f(x)) + f(0)</imath>. Hence, <imath>ax^{4} + bx^{2} = af(x)^{2} + bf(x) + f(0)</imath>. 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>
+<imath>x^{4} (a - a^{3}) - 2a^{2} b x^{3} - abx^{2} (b + 1) = f(0) = constant</imath>
-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 <imath>x</imath> terms must cancel out for that to be a constant. This can be achieved by either <imath>a = 0</imath> and <imath>b</imath> can be anything OR <imath>b = 0</imath> and <imath>a - a^{3} = 0</imath>. Now if <imath>a = 0</imath>, then <imath>f(x) = bx</imath> but recall that <imath>f(x)</imath> is an even function and hence <imath>b = 0</imath> is the only suitable option for a linear function to turn into an even function. Hence, <imath>f(x) = 0</imath> is a solution. If <imath>b = 0</imath> and <imath>a - a^{3} = 0</imath>, then <imath>a = 1, -1, 0</imath> and hence <imath>f(x) = x^{2}, -x^{2}</imath> 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 <imath>f(x)</imath> is not a polynomial.
-Case 2: <math>f</math> is not an injective function
+<b>Claim:</b> <imath>f</imath> has to be a polynomial to satisfy the given problem statement
-We can prove that <math>f</math> has to be injective. This is shown in sml1809's solution above in which we claim <math>f(a) = f(b) \neq 0</math> and substitute <math>x = a, y = b^{2}</math> and <math>x = b, y = a^{2}</math> to prove the claim that <math>f</math> has to be injective.
+<b>Proof:</b>
+We will first show that <imath>f(f(x)) = x^{2} f(x)</imath>. Substitute <imath>y = \frac{x^{2}}{2}</imath>:
+<imath>f(\frac{x^{2}}{2}) + x^{2} f(x) = f(f(x)) + f(\frac{x^{2}}{2})</imath>. Hence, the lemma. <imath>\square</imath>
+We have already shown that if <imath>f</imath> is injective, then <imath>f</imath> is even and this proof lies in the case of when <imath>f</imath> is injective. Hence, <imath>f(x)</imath> is an even function. Recall that <imath>f(x^{2}) = f(f(x)) + f(0)</imath>. Hence, <imath>f(x^{2}) = x^{2} f(x) + f(0)</imath>. Now we plug in <imath>x = 1</imath> to get <imath>f(1) = f(1) + f(0)</imath> and hence <imath>f(0) = 0</imath>. In other words, <imath>f(x^{2}) = x^{2} f(x)</imath>. (Haven't finished proof yet)
+Case 2: <imath>f</imath> is not an injective function
+We can prove that <imath>f</imath> has to be injective. This is shown in sml1809's solution above in which we claim <imath>f(a) = f(b) \neq 0</imath> and substitute <imath>x = a, y = b^{2}</imath> and <imath>x = b, y = a^{2}</imath> to prove the claim that <imath>f</imath> has to be injective.
+~ilikemath247365
+==Solution 4(Another solution)==
+We will show <imath>x^2 f(x) = f(f(x))</imath>. First, the original equation is
+<imath>f(x^2 - y) + 2yf(x) = f(f(x)) + f(y)</imath>
+We will make the substitution <imath>y = x^2 - 1</imath>
+<imath>f(1) + 2(x^2 - 1)f(x) = f(f(x)) + f(x^2 - 1)</imath>
+We will now take the original equation again and substitute <imath>y = 1</imath>
+<imath>f(x^2 - 1) + 2f(x) = f(f(x)) + f(1) \implies f(x^2 - 1) = f(f(x)) + f(1) - 2f(x)</imath>
+Now we will substitute this new representation of <imath>f(x^2 - 1)</imath> into our equation to get
+<imath>f(1) + (2x^2 - 2)f(x) = f(f(x) + f(f(x)) + f(1) - 2f(x) \implies (2x^2)f(x) = 2f(f(x)) \implies x^2 f(x) = f(f(x)) \square.</imath>
+We will now plug this in to our given equation
+<imath>f(x^2 - y) + 2yf(x) = f(f(x)) + f(y) \implies f(x^2 - y) + 2yf(x) = x^2 f(x) + f(y) \implies f(x^2 - y) - f(y) = f(x)(x^2 - 2y)</imath>.
+Now we will make the substitution <imath>x = y</imath>
+<imath>f(x^2 - x) - f(x) = f(x)(x^2 - 2x) \implies f(x^2 - x) = f(x)(x^2 - 2x + 1) \implies f(x(x - 1)) = f(x)(x - 1)^2</imath>. This immediately raises a conjecture. If we have two real numbers <imath>a, b</imath> where <imath>f</imath> is defined at these two real numbers, then
+<imath>f(ab) = f(a) b^2</imath>
+Let's prove this. From our equation <imath>f(x^2 - y) + 2yf(x) = x^2 f(x) + f(y)</imath>, we can substitute <imath>y = -x</imath> to get
+<imath>f(x^2 + x) - 2xf(x) = x^2 f(x) + f(-x) \implies f(x^2 + x) = f(x)(x^2 + 2x) + f(-x)</imath>. If we can show <imath>f(-x) = f(x)</imath>, then <imath>f(x^2 + x) = f(x(x + 1)) = f(x)(x^2 + 2x + 1) = f(x) (x + 1)^2 \implies f(x(x + 1) = f(x) (x + 1)^2</imath> and we should be done as we can now prove <imath>f(ab) = f(a) b^2</imath> by induction. So let's prove <imath>f(-x) = f(x)</imath>.
+The proof is simple. From our given equation <imath>f(x^2 - y) + 2yf(x) = f(f(x)) + f(y)</imath>, let's substitute <imath>x = 0</imath> to get
+<imath>f(-y) = f(f(0)) + f(y)</imath>.
+We will prove <imath>f(0) = 0</imath>. From our given equation <imath>f(x^2 - y) + 2yf(x) = f(f(x)) + f(y)</imath>, we will plug in <imath>x = y = 1</imath> to get
+<imath>f(0) + 2f(1) = f(f(1)) + f(1) \implies f(0) + f(1) = f(f(1)) \implies f(0) + f(1) = 1^2 f(1) \implies f(0) = 0</imath> following <imath>f(f(x)) = x^2 f(x)</imath>. Henceforth, we go back to <imath>f(-y) = f(f(0)) + f(y) \implies f(-y) = f(0) + f(y) \implies f(-y) = f(y) \implies f(-x) = f(x) \square.</imath>
+Hence, as <imath>f(-x) = f(x)</imath>, we had concluded before that if this was the case,
+<imath>f(x(x + 1)) = f(x) (x + 1)^2</imath> and <imath>f(x(x - 1)) = f(x) (x - 1)^2</imath> where the second equation we had derived before proving <imath>f(-x) = f(x)</imath>. From these two equations, we can generate any two real numbers that differ by <imath>1</imath> where
+<imath>f(ab) = f(a) b^2</imath> where <imath>a - b = 1</imath>.
+We will show this happens for all real <imath>a, b</imath> as we have proved only part of our conjecture. Note that <imath>f(a) = f(-a)</imath> has been proved.
+We will assume <imath>f(ab) = f(a) b^2</imath> only holds true if and only if <imath>a - b = 1</imath>. Now let <imath>b = -1</imath>. We have
+<imath>f(-a) = f(a)</imath> which holds true for all real numbers. However, we assumed the above was true if and only if <imath>a - b = 1</imath>. This would imply the above was true if and only if <imath>a = 0</imath> which is not the case. Hence, this is a contradiction and we have proved our conjecture. <imath>\square.</imath>
+Now, we have shown <imath>f(ab) = f(a) b^2</imath> is true for all real <imath>a, b</imath>. Let <imath>a = 1</imath> to get
+<imath>f(b) = b^2</imath> which means <imath>f(x) = x^2</imath> is a solution.
+Now consider <imath>ab = x \implies b = \frac{x}{a}</imath>. Now <imath>f(x) = f(a) \frac{x^2}{a^2} \implies \frac{f(x)}{x^2} = \frac{f(a)}{a^2}</imath>. From here we can generate many quadratic functions. Knowing <imath>f(x) = x^2</imath> is a solution gives us motivation for <imath>f(x) = -x^2</imath> which we can confirm will satisfy the given equation. We can prove this also
+Given the original equation <imath>f(x^2 - y) + 2yf(x) = f(f(x)) + f(y)</imath>, we have found <imath>f(x) = x^2</imath> is a solution. We will show that if <imath>f(x)</imath> is a solution, so is <imath>-f(x)</imath>. First, we'll assume something different, then prove by contradiction. We claim that if <imath>f(x)</imath> is a solution, <imath>-f(x)</imath> is not a solution. Let's plug in <imath>f(x) = x^2</imath> since we already know that's a solution. We have
+<imath>(x^2 - y)^2 + 2x^2 y = x^4 + y^2 \implies x^4 - 2x^2 y + y^2 + 2x^2 y = x^4 + y^2</imath> which holds.
+So now since we know <imath>f(x) = x^2</imath> is a solution, we had claimed that <imath>-f(x) = -x^2</imath> was not a solution. Hence we can plug this in to get
+<imath>-(x^2 - y)^2 - 2x^2 y = f(-x^2) - y^2 \implies -(x^2 - y)^2 - 2x^2 y = -x^4 - y^2</imath> of which this is exactly the negation of the equation we got for <imath>f(x) = x^2</imath>. Therefore <imath>-f(x)</imath> is indeed a solution and we have a contradiction with <imath>-f(x)</imath> not being a solution. Thus <imath>-f(x)</imath> has to be a solution. <imath>\square</imath>.
+Now assuming <imath>f</imath> is constant, we have
+<imath>f(x) = c \implies c + 2yc = x^2 c + c \implies 2yc = x^2 c \implies c = 0</imath> is a solution. Hence <imath>f(x) = 0</imath> is also a solution and so <imath>f(x) = x^2, -x^2, 0</imath>
 ~ilikemath247365

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

Page

Toolbox

Search