mihajlon wrote: Let $A(x,y)$ be assertion of $(f(x)+xy)\cdot f(x-3y)+(f(y)+xy)\cdot f(3x-y)=(f(x+y))^2$. If $A(0,0)\Longrightarrow 2(f(0))^2=(f(0))^2\Longrightarrow$ since $2\neq 1$ then $f(0)=0$. Let $A(0,y)\Longrightarrow (f(0))^2+f(y)\cdot f(-y)=f(y)\cdot f(-y)=(f(y))^2\Longrightarrow f(x)=0 \ \forall x\in \mathbb{R}$ or $f(y)=f(-y)... \bigstar$. Now let $A(x,-x)\Longrightarrow (f(x)-x^2)\cdot f(4x)+(f(-x)-x^2)\cdot f(4x)=(f(0))^2\stackrel{\bigstar}{\Longleftrightarrow} 2\cdot (f(x)-x^2)\cdot f(4x)=0$. So we have two possibilities: $\textbf{1.)}$ $f(x)=x^2, \ \forall x\in \mathbb{R}$, by checking i.e. $(x^2+xy)(x-3y)^2+(y^2+xy)(3x-y)^2=(x+y)^4$, we get it's true. $\textbf{2.)}$ $f(4x)=0$. Since this hold for all $x\in \mathbb{R}$ we get $f(x)=0$. Thus only solutions are $\boxed{f(x)=x^2; \ f(x)=0}, \ \forall x\in \mathbb{R}$ This solution doesn't show that $f(x) = 0$ or that $f(x) = x^2$ is true for all real numbers. It just shows that this is true for each one. I made the same mistake for the 2014 USAMO.

pointwise value trap :O

Hmm this problem was not nice. It took me 4 hours to come up with the correct argument to avoid pointwise value.

I could've sworn that this and 2014 AMO2/JMO3 were the same problem.

My solution outline: 1.Prove that f(0)=0 2.Prove that the function is even 3.Prove that for all x f(x)=x^2 or f(4x)=0 4.Prove that for all x f(x)=x^2 or f(x)=0 5.Using analysis to prove that limit of the ratio of LHS to RHS is 3/4 and arrive at a contradiction, if f(x)=0 and f(y)=y^2 for some x,y not equal to 0.

lucylai wrote: This solution doesn't show that $f(x) = 0$ or that $f(x) = x^2$ is true for all real numbers. It just shows that this is true for each one. I made the same mistake for the 2014 USAMO. I'm little bit confused, how does not simple checking show it's true for all reals?(I'm unexperienced...)

I'm not sure if my solution is valid, but... I first proved that $f(0) = 0$ and I proved that $f(x) = 0$ satisfied the functional equation (I just plugged it in the functional equation), and then I proved that it is even. Then I proved that if it is not $f(x) = 0$ then it has to be $x^2 + cx$ for some $c$ and then said that it was even so it has to be $f(x) = 0$ or $f(x) = x^2$.

Showed that if f(t)=0, then f(nt/2^m)=0 for all n, m positive integers. Then I said as m tends to infinity, this thing gets dense, and now I don't know if this is 7- or 0+

@aopser123 Those functions do work for all reals, but you didn't show that a function with $f(r)=0$ and $f(s)=s^2$ does not work.

Well actually I tried to prove that the function should be continuous for an hour but got nothing so I wrote "it is easy to see that..."... Probably 2 pts deduction?

^ probably going to be like a 5 point deduction.

ok I will post my sketch here as well Eventually one gets $f(x)=x^2$ or $f(x)=0$ for each $x$. Suppose for some nonzero $j,k$ we had $f(k)=k^2$ and $f(j)=0$. Assume they are positive because $f$ is even. Plugging in $(x,y)=(j,-j)$ implies that $f(4j)=0$. Now $f(4^m j)=0$, so we can assume $j$ is arbitralily large. Find suitable $x$ and $y$ so that $x-3y=k$ and $3x-y=j$. Then we have $(f(x)+xy)k^2=f(x+y)^2$. The LHS is clearly positive (taking $j$ large enough makes $x$ positive), so $f(x+y)^2=(x+y)^4$. The LHS is at most $x(x+y)k^2$. Solving for $x$ and $y$ in terms of $j$ and $k$, and plugging in the results, it is easy to see that if $j$ is large enough a contradiction will arise.

WL0410 wrote: @aopser123 Those functions do work for all reals, but you didn't show that a function with $f(r)=0$ and $f(s)=s^2$ does not work. Wgat are $r$ and $s$?

i showed $f(4x)=0$ or $f(x)+f(-x)=2x^2$ then piecewise just confused me to death

This problem was quite ugly. My solution worked by a miracle

Steps 5 and 6 of the above solution are pretty obscure, so I wanted to address the motivation for them. We have already proved that $f(t) = 0$ or $f(t) = t^2$ for each $t$, and would now like to show that if $f \not\equiv 0$, then $f(t) = 0 \implies t = 0.$ In some sense, we have $2^5$ cases according to the five appearances of $f$ in the given equation. Now, we know that setting $f(t) = t^2$ for each of these appearances works, in which case the equation reads $\left(x + y\right)^4 = \left(x + y\right)^4.$ Hence, if we assume that $f(x + y) = 0$ on the RHS, and can force $f(t) = t^2$ in the four applications of $f$ on the LHS, then we will arrive at the equation $\left(x + y\right)^4 = 0$, which forces $x + y = 0$, as desired. This reasoning motivates Step 5, from which Step 6 follows after playing around a bit.

niraekjs wrote: ^ probably going to be like a 5 point deduction. I do fear so. https://www.artofproblemsolving.com/wiki/index.php?title=2014_USAMO_Problems/Problem_2

jeff10 wrote: Then I said as m tends to infinity, this thing gets dense, and now I don't know if this is 7- or 0+ I pretty much said the same thing, "because there are large enough values of $x$ so that $f(x) = 0$ we get a contradiction somewhere." Don't know how many points I'm getting. What happened to a move away from trash #4's -_-

The only solutions are $f\equiv 0$ and $f\equiv x^2$, it is easy to check both work. We now show these are the only solutions. Take $x=y=0$ to yield $f(0)^2+f(0)^2=f(0)^2$, hence $f(0)=0$. Claim: $f$ is even. Solution: Take $x=0$ to yield $f(y)f(-y)=f(y)^2$. We similarly write $f(y)f(-y)=f(-y)^2$. If $f(y)=0$, we have $f(-y)=0$ by the latter piece of information. Otherwise, the former piece of information gives $f(-y)=f(y)$. $\fbox{}$ Claim: If $f(t)\ne 0$, then $f(2t)\ne 0$. Solution: Take $t=4x$ and $y=3x$ to yield \[(f(x)+3x^2)\cdot f(2t) = (f(x)+3x^2)\cdot f(8x) = f(4x)^2 = f(t)^2,\]implying the desired. $\fbox{}$ Take $x=-y$ to yield \[(f(x)+f(-x)-2x^2)f(4x)=(f(x)-x^2)f(4x)+(f(-x)-x^2)f(4x)=0.\]If $f(4x)\ne 0$, this yields \[2x^2=f(x)+f(-x)=2f(x)\implies f(x)=x^2,\]so we have extracted the initial pointwise trap; for each $x$, we have $f(x)=x^2$ if $f(4x)\ne 0$. We use the following argument to reduce the pointwise trap to a reasonable one. Take $x=y=t$ to yield \[(f(t)+t^2)\cdot f(-2t)+(f(t)+t^2)\cdot f(2t)=f(2t)^2.\]Given that $f(2t)\ne 0$, this yields \[f(2t)=f(t)+t^2+f(t)+t^2=2(f(t)+t^2)\qquad (\heartsuit).\]Now, for $x$ such that $f(4x)\ne 0$, we determine that $f(x)\ne 0$ (as $x\ne 0$), so $f(2x)\ne 0$. Invoking $\heartsuit$ twice, we find $f(2x)=4x^2$ and $f(4x)=16x^2$, so if $t=4x$, we now have that either $f(t)=0$ or $f(t)=t^2$. Remark that since $f(2t)$ cannot equal $2t^2$ for $t\ne 0$, $f(2t)\ne 0$ implies $f(t)\ne 0$, so our earlier claim may be strengthened to ``$f(t)\ne 0$ iff $f(2t)\ne 0$.'' We also remark that $f(t)\ge 0$ for all $t$ by the pointwise trap. Now, suppose there exist $a,b\ne 0$ for which $f(a)=0$ and $f(b)\ne 0$. Since $f$ is even, we may assume that $a,b>0$. Note we may pick $k$ for which $2^ka>b$, so we can assume $a>b$. Pick $x,y>0$ for which $x-3y=b,x+y=a$. By the given on $x,y$, we have \[(f(x)+xy)\cdot f(b)+(f(y)+xy)\cdot f(3x-y)=0.\]Since $f$ is nonnegative, this implies $f(b)=0$ and we have a contradiction. Hence, either $f$ is identically zero or it is $x^2$, as desired.

Exhausting problem, but pretty cool. The answer is $f(x)=x^2$ and $f \equiv 0$, which we can check works. We will denote the assertion as $P(x,y)$. First from $P(0,0)$ we get: $$f(0)^2+f(0)^2=f(0)^2$$from which we derive $f(0)=0$. Next, from $P(0,y)$ we get: $$f(y)f(-y)=f(y)^2.$$From $P(0,-y)$ we also get: $$f(y)f(-y)=f(-y)^2.$$We claim that this implies $f$ is even. Indeed, if for some $y$ we have $f(y) \neq 0$ then we can divide both sides of the first equation to get $f(y)=f(-y)$. Otherwise, if $f(y)=0$, the second equation gives $f(-y)=0$ as well, which also means $f(y)=f(-y)$ as desired. Now from $P(x,-x)$ and using $f(0)=0$ with $f$ even, we get: $$f(4x)(2f(x)-2x^2)=0$$which gives $f(x)=x^2$ or $f(4x)=0$ for all $x \in \mathbb{R}$. This means that $f(x)=0$ for some $x$ implies $f(4x)=0$ (if $x=0$ the statement is vacuous, else we are forced to have $x^2=0$ which is a contradiction). Now, suppose that for some $a$ we have $f(4a)=0$. We claim that this implies $f(a)=0$ as well. From $P(x,3x)$ with $x$ arbitrary we get: $$(f(x)+3x^2)f(8x)=f(4x)^2$$Now letting $x=\tfrac{a}{2}$ gives $f(2a)=0$, and then letting $x=\tfrac{a}{4}$ gives $f(a)=0$. Thus for all $x$, we either have $f(x)=x^2$ or $f(x)=0$ (which implies $f(4x)=0$ as well). We also have that: $$f(x)=0 \iff f(4x)=0$$for all $x$. This implies $f(x) \geq 0$ for all $x$. Suppose now that there exists some value of $a \neq 0$ such that $f(a)=0$. Since $f$ is even, WLOG let $a$ be positive. We will show that $f \equiv 0$. By $P(a,-y)$ for $y \in (0,a)$ and rearranging, we get: $$-ayf(a+3y)+(f(y)-ay)f(3a+y)-f(a-y)^2=0$$But on the other hand, we have: $$-ayf(a+3y) \leq 0$$and: $$(f(y)-ay)f(3a+y) \leq y(y-a)f(3a+y) \leq 0.$$which means that unless $f(a-y)^2=0$, the $LHS$ will be negative. This implies that $f(a-y)=0$ for all $y \in (0,a)$ which implies $f(x)=0$ for all $x \in (0,a)$. We claim that this result, combined with $f(x)=0 \iff f(4x)=0$, is enough to imply that $f(x)=0$ for all positive $x$. Indeed, using easy induction and the previous result we get that $f(x)=0$ for all $x \in (0,4^ka)$ for some nonnegative integer $k$. This means that for any positive $b>a$, we can simply take a $k$ such that $4^ka>b$ (which clearly exists as $a>0$), and get that $f(b)=0$. Since $b$ is arbitrary, putting this with the previous result, we get that if there exists some $a \neq 0$ such that $f(a)=0$, then we must have $f(x)=0$ for all $x \geq 0$. But since $f$ is even, this means that $f \equiv 0$ everywhere. This therefore eliminates all the pathological solutions, leaving us only with $f \equiv 0$ and $f(x)=x^2$, as desired. $\blacksquare$ Edit: wait the pointwise step is wrong, but I have to go to bed now. In the process of fixing, the pointwise solution has completely changed and made much quicker. For the sake of completeness I'll also fix the "original", which will be hidden below:

Solved with jacoporizzo. Find all functions $f:\mathbb{R}\rightarrow \mathbb{R}$ such that for all real numbers $x$ and $y$, $$(f(x)+xy)\cdot f(x-3y)+(f(y)+xy)\cdot f(3x-y)=(f(x+y))^2.$$ The solutions are $f(x)=0$ and $f(x)=x^2$ only. Clearly, the first solution works. The second also does since \begin{align*} (x^2+xy)\cdot (x^2-6xy+&9y^2)+(y^2+xy)\cdot (9x^2-6xy+y^2)\\ &=x^4-6x^3y+9x^2y^2+x^3y-6x^2y^2+9xy^3\\ &\qquad +9x^2y^2-6xy^3+y^4+9x^3y-6x^2y^2+xy^3\\ &=x^4+4x^3y+6x^2y^2+4xy^3+y^4\\ &=(x+y)^4 \end{align*}for all $x$, $y$. Now, we show that these are the only solutions. Plug in $(x,y)=(0,0)$ to get $2f(0)^2=f(0)^2\implies f(0)=0$. Plug in $(x,y)=(3i, i)$ to get $0 + (f(i)+3i^2)\cdot f(8i)=f(4i)^2$. Plug in $(x,y)=(i,3i)$ to get $(f(i)+3i^2)\cdot f(-8i)+0=f(4i)^2$. These two give that $f(8i)=f(-8i)$ for all $i\neq 0$, so $f(x)=f(-x)$ for each $x$. Now, plugging $y=-x$, $$(f(x)-x^2)\cdot f(4x)+(f(-x)-x^2)\cdot f(4x) = 0.$$Hence, for each $x$, either $f(x)=x^2$ or $f(4x)=0$. Notice that this implies $f$ is nonnegative. $f(x)=x^2$ is one such solution. Now, assume that there exists some nonzero $k$ such that $f(k)=0$. Plug in $(x,y)=(k,k)$. Then, \[k^2\cdot f(2k)+k^2\cdot f(2k) = f(2k)^2\]which implies $f(2k)=0$ as well, since $k$ is nonzero. Finally, we plug in $y=k-x$. This gives \[(f(x)+x(k-x))\cdot f(4x-3k) + (f(k-x)+x(k-x))\cdot f(4x-k) = 0\] Notice that when $x\in (0,k)$, we have that all of the terms on the LHS are nonnegative and $x(k-x)$ is positive. Since the RHS is zero, each of the two terms on the LHS are zero. In particular, $f(4x-k)=0\implies f(i)=0$ for all $-k<i<3k$. From here, notice that we can "double" this interval repeatedly to get that $f(x)=0$ for all positive $x$. Then, since $f$ is even, $f(x)=0$ for all negative $x$ as well. This finishes.

The possible are $f \equiv 0$ and $f \equiv x^2$ Let $F(x, y)$ denote the given assertation. Firstly, we compute $f(0)$ by asserting $F(0, 0)$. We have $2 \cdot f(0)^2 = f(0)^2$, giving that $f(0) = 0$. Claim. $f$ is even. Proof. We assert $F(0, y)$ and $F(0, -y)$.$$F(0, y) \to f(y)f(-y) = f(y)^2$$$$F(0, -y) \to f(-y)(f(y) = f(-y)^2$$Hence, $f(-y)^2 = f(y)^2$. Suppose that for a certain $y$, we have $f(y) \ne f(-y)$. We have $f(-y) = -f(y)$, meaning that $f(y)f(-y) = -f(y)^2$, contradiction. Hence, $f(y) = f(-y)$ for all $y$ as desired $\square$ Now assert $F(x, -x)$, noting that we may substitute $f(3y - x)$ for $f(x - 3y)$. We have $f(4x)(f(x) - x^2) = 0$ for all real $x$, which implies that $f$ is non-negative. To finish, assume that for some $c$, $f(c) = 0$. Asserting $F(c, c)$ gives that $f(c) = 0 \to f(2c) = 0$. Notice we can scale up arbitrarily many times as needed and use the fact that $f$ is even to achieve a certain real from a real in the interval $(0, d)$, or $(d, 0)$. Hence, if for all reals $\in (0, d)$ or $(d, 0)$ that $f \equiv 0$, we have that $f \equiv 0$ across all reals for any real $d$,. Now assert $F(a, c - a)$, where $a \in (0, c)$. We have $$[f(a) + a(c - a)][f(4a - 3c)] + [f(c - a) + a(c - a)][f(4a - c)] = 0$$Since $a(c - a) > 0$, we have that both terms on the left hand side are at least zero, but since the right hand side is zero, we must have that both terms are zero. Since $f$ is non-negative, $f(i) + a(c - a) > 0$ for all real $i$, meaning that we must have $f(4a - 3c) = f(4c - a) = 0$. Taking $f(4a - c)$ yields that for all reals in the interval $(-c, 3c)$ $f$ is zero, meaning that for all reals in the interval $(0, 3c)$ as desired. Hence, if such a $c$ exists, $f(x) = 0$< and otherwise we must have $f(x) - x^2 = 0$, meaning that $f(x) = x^2$ as desired. Remarks. It took forever to get the last part. My solution ended up looking very similar to @above's but oh well. This was very exhausting but worth it.

We claim that the answers are $f(x)=0$ and $f(x)=x^2$ Set $(x,y)=(0,0)$ gives $f(0)=0$ Set $x=0$ gives $f(y)f(-y)=f(y)^2$ Which implies that $f$ is an even function Set $y=-x$ gives $f(4x)(f(x)-x^2)=0$. Set $y=3x$ gives $(f(x)+3x^2)f(8x)=f(4x)^2$ which implies that $f(x)=0\Longleftrightarrow f(2x)=0$. Therefore, $f(x) = \begin{cases} x^2 & ?? \\ 0 & ?? \end{cases}$ Let $f(a)=0$ and $f(b)=b^2$ for some $a,b>0$ (since $f$ is odd). So, there exist $a'=2^ka>b$, Let $a'=m+n$ and $b=m-3n$ Set $(x,y)=(m,n)$ gives $f(m)+f(n)+2mn=0$ Which is a contradiction since $m,n>0$

I claim that the only solutions are $f(x)=0$ and $f(x)=x^2$. These clearly work. Let $P(x,y)$ denote the given assertion. $P(0,0): f(0)^2=2f(0)^2$ so $f(0)=0$. $P(0,y): f(y)f(-y)=f(y)^2$. So the function is even. $P(3y,y): (f(y)+3y^2)\cdot f(8y)=f(4y)^2$. Now if $f(8y)=0$, then $f(4y)=0$, so if $f(k)=0$, then $f(\frac{k}{2})=0$. $P(x,-x): (f(x)-x^2)\cdot f(4x)+(f(x)-x^2)\cdot f(4x)=2((f(x)-x^2)\cdot f(4x))=0$ Either $f(x)=x^2$ or $f(4x)=0 \implies f(2x)=0 \implies f(x)=0$. Now it remains to prove that we can't have $f(x)=0$ for some reals and $f(x)=x^2$ for other reals in the same function, excluding $0$. Suppose that $f(x+y)=(x+y)^2$ where $x$ and $y$ are both positive. We have\[(f(x)+xy)\cdot f(x-3y)+(f(y)+xy)\cdot f(3x-y)=(x+y)^4.\]The maximum of the LHS is $(x^2+xy)\cdot(x-3y)^2+(y^2+xy)\cdot (3x-y)^2=(x+y)^4$ (because $f(x)\ge0$ and $xy>0$), which only holds when $f(x)=x^2, f(y)=y^2, f(x-3y)=(x-3y)^2,$ and $f(3x-y)=(3x-y)^2$. Call $x$ good if $f(x)=x^2$. Call $x$ bad if $f(x)=0$. We have all real values $k$ such that $0\le k \le x+y$ are good because we can write $x+y$ as $k+l$, where both $k$ and $l$ are positive. Also, if $k=0$ or $x+y$, $k$ is obviously good. Let there be positive reals $m$ and $n$ such that $m+n$ is bad. If we have some good $a$, we can multiply $a$ by a power of $2$ so that we exceed $m+n$, which implies $m+n$ is good, a contradiction (if $a\cdot$ (the power of $2$) is bad, then $a$ is bad, so $a\cdot$ (the power of $2$) is good). So if we have a good positive number (by evenness if we have $-x$ is good, then $x$ is good), then all positive numbers are good, and because of evenness, all reals are good. Thus, all reals are good or all reals are bad. So we are done.

This was very fun to write-up Find all functions $f:\mathbb{R}\rightarrow \mathbb{R}$ such that for all real numbers $x$ and $y$, $$(f(x)+xy)\cdot f(x-3y)+(f(y)+xy)\cdot f(3x-y)=(f(x+y))^2.$$ Let $P(x, y)$ be the assertion that $(f(x)+xy)\cdot f(x-3y)+(f(y)+xy)\cdot f(3x-y)=(f(x+y))^2.$ Then, $P(0, 0)$ yields $2f(0)^2 = f(0)^2 \rightarrow f(0)=0$. Claim 1: $f$ is even. Proof: $P(0,y)$ yields $f(y)f(-y)=f(y)^2$. Similarly, $P(0, -y)$ yields $f(-y)f(y)=f(-y)^2$. It follows from here that $f$ is even. Claim 2: For each $x \in \mathbb{R}$, we have that $f(x)=0$ or $f(x) = x^2$. Proof: $P(x, -x) \rightarrow (f(x)-x^2) \cdot f(4x) + (f(-x)-x^2) \cdot f(-4x) = (f(0))^2 \rightarrow (f(x)-x^2) \cdot f(4x) = 0$. So, either $f(x)=x^2$ or $f(4x) = 0$. $P(x, 3x) \rightarrow (f(x) + 3x^2) \cdot f(8x) = (f(4x))^2$. This means that $f(8x)=0 \implies f(4x)=0$, which is equivalent to $f(2x)=0 \implies f(x)=0$. Therefore, $f(4x) = 0 \implies f(x)=0$. Thus, $f(x)=0$ or $f(x) = x^2$. Now, assume that the function is $0$ at some nonzero value. This implies the existence of a positive $C$ such that $f(C) = 0$. Now, for any $0 < D \le 2C$, the solutions to the equations $$ \begin{cases} x+y = C \\ 3x-y = D \end{cases} \rightarrow \begin{cases} x = \frac{C+D}{4} \\ y = \frac{3C-D}{4}. \end{cases} $$are always strictly positive. Then $$P(x, y) \rightarrow (f(x)+xy)\cdot f(x-3y)+(f(y)+xy)\cdot f(3x-y)=(f(x+y))^2.$$Since the RHS is $0$, and $(f(x)+xy)\cdot f(x-3y)$ is nonnegative and $(f(y)+xy)$ is strictly positive, we have that $f(3x-y) = f(D)$ is also $0$, for all $0 < D \le 2C$. A simple induction argument can show that the function is $0$ for all real values. Therefore, the only possible functions are $\boxed{f(x)=x^2 \text{ or } f(x)=0}$ and we can indeed check that it works.

Let $P(x,y)$ be the given assertion. $P(0,0)$ gives $f(0)=0.$ And $P(0,x)$ yields if $f(x)\neq 0$ then $f(x)=f(-x),$ although it works for $f(x)=0$ as well. From $P(x,-x)$ we get $f(x)=x^2$ or $f(4x)=0.$ Assume $f(4x)=0,$ then $P(3x/4,x/4)$ gives $f(2x)=0.$ Reapplication gives $f(x)=0.$ Claim: $\exists x\neq 0: f(x)=0 \implies f\equiv 0.$ Proof. WLOG $a,b >0: f(a)=0$ and $f(b)\geq 0.$ But $P(a/2,-b/2)$ forces $f(b)=0.$ $\blacksquare$ So the solutions are $f\equiv \text{Id}^2$ and $f\equiv 0.$ Both work.

$(f(x)+xy)f(x-3y)+(f(y)+xy)f(3x-y)=f(x+y)^2$ Let $P(x,y)$ be assertion of this equation. 1.$P(0,0)$ $f(0)=0$ 2.$P(x,x)$ $(f(x)+x^2)(f(-2x)+f(2x))=f(2x)^2$ 3.$P(x,-x)$ $f(4x)(f(x)+f(-x)-2x^2)=f(0)=0$ So,there are two cases which we should consider. 1'.$f(4x)=0$ which means $f$ is const and $f(x)=0$ and it indeed fits. 2'.$\boxed{f(x)+f(-x)=2x^2}$ 4.$P(2x,-2x)$ $f(2x)+f(-2x)=8x^2$ $\implies{(f(x)+x^2).8x^2=f(2x)^2}$ 5.$P(-x,-x)$ $(f(-x)+x^2)(f(2x)+f(-2x))=f(-2x)^2$ $5.-4.$ $\implies{f(-x)-f(x)=f(-2x)-f(2x)}$ it is easy to see $f(x)=f(2x)-3x^2$ Let's put this fact on 4. then we get $f(x).2x^2=x^4+f(2x)^2$ which means $(x^2-f(x))^2=0$ thus we have $\boxed{f(x)=x^2}$ Only thing check pointwise trap.

Only solution is $f(x)=0\forall x$ or $f(x)=x^2\forall x$. Take $x=y=0$ gives $f(0)=0$. Then take $x=0$ and use that $f(0)=0$, giving $f(y)\cdot f(-y)=f(y)^2$ So $f$ is a even function or always $0$, which is still even anyways. Take $x=-y$ and use that $f$ is even, $f(0)=0$, we can get that $$2(f(y)-y^2)f(4y)=0$$This gives that for any $x$, we must have $f(x)=x^2$ or $f(x)=0$. The solution set follows after we deal with the pointwise trap. Full proof here: https://infinityintegral.substack.com/p/usajmo-2016-contest-review

plug $y=0$ to get $f(0)f(3x)=0$, so $f(0)=0$ or all $f(x)=0$ plug $x=0$ to get $f(y)f(-y)=(f(y))^2$ and $f(-y)=f(y)$ plug $y=-x$ to get $(f(x)+f(-x)-2x^2)(f(4x))=0$ either all $f(x)=0$, or $f(x)+f(-x)=2x^2$ $f(x)=f(-x)$, so solving $f(x)+f(-x)=2x^2$ gives $f(x)=x^2$ therefore, our only solutions are $f(x)=0$ or $f(x)=x^2$

First, note that plugging in $(0,0)$ gives that \[f(0)^2+f(0)^2=f(0)^2,\]which just implies that $f(0)=0$. Using this, we plug in $(0,y)$ to get the equation \[(f(0)+0)f(-3y)+(f(y)+0)f(-y)=f(y)^2 \iff f(y)f(-y)=f(y)^2,\]and plugging in $(0,-y)$ similarly yields that $f(y)f(-y)=f(y)^2$, meaning that either $f(y)=f(-y)\neq 0$, or $f(y)=f(-y)=0$. Either way, we have that $f(y)=f(-y)$. Now, we plug in $(x,-x)$ to get that \[(f(x)-x^2)f(4x)+(f(-x)-x^2)f(4x)=f(0)^2 \iff (f(x)-x^2)f(4x)=0,\]which implies that either $f(x)=x^2$ or $f(4x)=0$. Replacing $x$ with $\frac{x}{4}$ yields that for all $x$, $f(x)$ only has two possible values - $0$ and $x^2$. Now we are simply left with proving that there is no pointwise trap. I will prove this by proving that if $\exists a\neq 0$ such that $f(a)=0$, then $f(x)=0$ for all real numbers $x$. From this we can conclude that otherwise, $f(x)=x^2$ for all $x\in \mathbb{R}$. Assume FTSOC that $\exists a\neq 0$ such that $f(a)=0$. C1: First, I claim that if $f(a)=0$, then we also have that $f(2a)=f\left(\frac{a}{2}\right)=0$. The latter can be shown by plugging in $\left(\frac{3a}{4}, \frac{a}{4}\right)$, which gives \[\left(f\left(\frac{3a}{4}\right)+3\left(\frac{a}{4}\right)^2\right)f(0)+\left(f\left(\frac{a}{4}\right)+3\left(\frac{a}{4}\right)^2\right)f(2a)=f(a)^2,\]\[\iff 0=f(a)^2=\left(f\left(\frac{a}{4}\right)+3\left(\frac{a}{4}\right)^2\right)f(2a),\]and since $f\left(\frac{a}{4}\right)$ is always $\geq 0$ (since $f(x)=0$ or $x^2$) and $a$ is non-zero, we have that $f\left(\frac{a}{4}\right)+3\left(\frac{a}{4}\right)^2$ must also be greater than zero. This gives that we must have $f(2a)=0$. Reverse engineering the process, we can also get that $f\left(\frac{a}{2}\right)=0$, similarly. C2: Now I claim that if there exist two nonzero numbers such that $f(x)=f(y)=0$, then $f(x-y)=0$. This can be proved by setting $a=\frac{3y-x}{8}$ and $b=\frac{y-3x}{8}$. Plugging in $(a,b)$ then gives that \[(f(a)+ab)f(x)+(f(b)+ab)f(y)=f(a+b)^2,\]\[\iff 0=f(a+b)^2,\]\[\iff f\left(\frac{x-y}{2}\right)=f\left(\frac{y-x}{2}\right)=f(a+b)=0,\]and by C1, we have that since $f\left(\frac{x-y}{2}\right)=0$, we have that $f(x-y)=0$, proving our claim. C3: Finally, I claim that if $f(x)=0$ is not universal for all real numbers $x$, then there cannot exist $a\in \mathbb{R}$ such that $a\neq 0$ and $f(a)=0$. Assume FTSOC that there exists such an $a$. Then let $x+y=a$ such that $f(x)\neq 0$. By our assumption, this $x$ always exists. First notice that since $f(x)\neq 0$, this implies that $f(y)\neq 0$ by C2. Similarly by C2, we also get that $f(3x-y)\neq 0$, since we have that if $f(3x-y)=0$, we'd get \[f(a)=f(x+y)=0,\]\[\iff f(a-(3x-y))=0,\]\[\iff f(4x)=0 \iff f(2x)=0 \iff f(x)=0,\]using C1, a contradiction. Therefore, $f(3x-y)\neq 0$. Similarly, we can also get that $f(x-3y)\neq 0$. Using the fact that for all real numbers $r$, we have that $f(r)=r^2$, if we plug this (meaning $(x,y)$) into the large equation, we'd get that \[(x^2+xy)(x-3y)^2+(y^2+xy)(3x-y)^2=f(x+y)^2=f(a)^2=0,\]\[\iff (x+y)(x)(x^2-6xy+9y^2)+(x+y)(y)(9x^2-6xy+y^2)=0,\]\[\iff (x+y)(x^3-6x^2y+9xy^2+9x^2y-6xy^2+y^3)=0,\]\[\iff (x+y)(x^3+3x^2y+3xy^2+y^3)=0,\]\[\iff (x+y)(x+y)^3=a^4=0,\]a contradiction, since we assumed that $a\neq 0$! Therefore, our only solutions to this FE are $f(x)=0$ for all real $x$ or $f(x)=x^2$ for all real $x$, finishing the problem.

Denote the assertion as $P(x,y)$ When $x=y=0, 2f(0)^2=f(0)^2, f(0)=0$ We have $P(0,x)\implies f(x)f(-x)=f(x)^2, f(-x)=f(x)$ or $f(x)=0; P(x,x)\implies (f(x)+x^2)f(-2x)+(f(x)+x^2)f(2x)=2f(2x)(f(x)+x^2)=f(2x)^2\implies f(2x)=2(f(x)+x^2)$ Assertion $P(x,3x)\implies (f(x)+3x^2)f(8x)=f(4x)^2$ By the fact that $f(2x)=2(f(x)+x^2)$, we can have $f(4x)=2(f(2x)+4x^2)=4f(x)+12x^2; f(8x)=2(f(4x)+16x^2)=8f(x)+56x^2$ Thus, we write $(f(x)+3x^2)f(8x)=f(4x)^2$ in terms of $f(x)$, we attain $f(x)^2+2x^2f(x)-3x^4=0$. We could solve $f(x)=x^2$ or $f(x)=-3x^2$. We have to get rid of the $f(x)=-3x^2$ since $P(1,1)$ doesn't work, $LHS=-48\neq 48=RHS$, while it is easy to check $f(x)=x^2$ works Thus, we have $\boxed{f(x)=0; f(x)=x^2}$ as solutions.

The solutions are $f(x) = 0$ and $f(x) = x^2$, both of which are somewhat easy to verify. Taking $P(0,0)$ gives us $f(0) = 0$ – now, adding $P(0,x)$ and $P(0,-x)$ gives us $(f(x)-f(-x))^2 = 0$, which implies that $f$ is even. Taking $P(x,-x)$ gives us \[f(4x)(f(x)-x^2) = 0, \qquad (\diamondsuit)\]while taking $P(x,x)$ gives us \[f(2x) (f(2x)-2f(x)-2x^2) = 0. \qquad (\heartsuit) \] Claim: We have $f(x) \in \{0, \tfrac{x^2}{2}, x^2 \}$ for all $x$. Proof: Assume that $x$ is a real which contradicts the claim. From $\heartsuit ( \tfrac{x}{2} )$, we know that $f(\tfrac{x}{2}) \not \in \{0, \tfrac{x^2}{4} \}$. Also, from $\diamondsuit( \tfrac{x}{4} )$, we know that $f( \tfrac{x}{4} ) = \tfrac{x^2}{16}$. But now, $\heartsuit (\tfrac{x}{4} )$ gives us \[f( \tfrac{x}{2} ) \left( f( \tfrac{x}{2} ) - \tfrac{x^2}{4} \right) = 0,\]which is a contradiction. Claim: If $f(C)= 0$ for some nonzero $C$, then $f(x) = 0$ for all $x$. Proof: Assume WLOG that $C$ is positive. Letting $x$ and $y$ be positive reals such that $x+y = C$, $P(x,y)$ implies that both $f(x-3y)$ and $f(3x-y)$ are zero, since we know $f$ is nonnegative now. Since $x-3y$ takes values in the range $(-3C, C)$, we can induct to show the claim.