OOps sorry for my stupid mistakes

Abdek wrote: $\Phi(x,x,z) \Rightarrow f(f(z)-2x^2)+2f(2xz)=f(f(z)+2x^2)+2f(0)$ let $x=\sqrt{\frac{\lambda}{2}}$ then from the last equality we get $ f(f(z)-\lambda)+2f(2\lambda z)=f(f(z)+\lambda)+2f(0)$ I don't think that's quite right (In this case $2xz\neq2\lambda z$. Indeed, $f(x)=x^2$ is another solution to this equation.

This problem has been discussed here.

zhero, your ultimate solution is very nice. I was first thinking about f(0)=0 and also f(x)=f(xf(1)) Nice Solution! Did you take the APMO Test and solve it?

Posted in the other thread as well but posting here for reference purposes. APMO 2010/5 wrote: Find all functions $f$ from the set $\mathbb{R}$ of real numbers into $\mathbb{R}$ which satisfy for all $x, y, z \in \mathbb{R}$ the identity $$f(f(x)+f(y)+f(z))=f(f(x)-f(y))+f(2xy+f(z))+2f(xz-yz).$$

I am not sure that $\mathbb{S}=R$ implies $f-$ constant. The function $f(2)=1,f(x)=2$ for $x\neq 2$ satisfies $\mathbb{S}=R,$ but is not constant.

Kirilbangachev wrote: I am not sure that $\mathbb{S}=R$ implies $f-$ constant. The function $f(2)=1,f(x)=2$ for $x\neq 2$ satisfies $\mathbb{S}=R,$ but is not constant. The claim "$f$ is constant" follows from the fact that $f(x)=f(xs)$ for all $s \in \mathcal{S}$ and $\mathcal{S}=\mathbb{R}$; not just the latter.

anantmudgal09 wrote: Kirilbangachev wrote: I am not sure that $\mathbb{S}=R$ implies $f-$ constant. The function $f(2)=1,f(x)=2$ for $x\neq 2$ satisfies $\mathbb{S}=R,$ but is not constant. The claim "$f$ is constant" follows from the fact that $f(x)=f(xs)$ for all $s \in \mathcal{S}$ and $\mathcal{S}=\mathbb{R}$; not just the latter. Yeah, my bad. Thanks

Let $P(x,y,z)$ be the statement $f(f(x)+f(y)+f(z))=f(f(x)-f(y))+f(2xy+f(z))+2f(xz-yz)$. Take the difference in $P(x,y,z)$ and $P(y,x,z)$, then we get $$f(f(x)-f(y))-f(f(y)-f(x))=2f((x-y)z)-2f((y-x)z)$$Note that $x-y$ represents all real number $t$, so for any $t\in \mathbb{R}$, $f(tz)-f(-tz)$ is constant when varying $z$. So $$f(t)-f(-t)=f(t(-1))-f((-t)(-1))=f(-t)-f(t)\Rightarrow f(t)=f(-t) \enspace \forall t\in \mathbb{R}$$So $f$ is an even function. Compare $P(x,y,0)$ with $P(x,-y,0)$ we get $f(2xy+f(0))=f(-2xy+f(0))=f(2xy-f(0))$, so let $t=2xy$ we see that $f(t)=f(t+2f(0))$ for all reals $t$. In psrticular, we also have $P(0,0,0)\Rightarrow f(3f(0))=f(f(0))+3f(0)$. Since $f(3f(0))=f(f(0))$, then $f(0)=0$. Now we prove that either $f$ is zero function, or $f(a)=f(b)\Rightarrow a=\pm b$. We give three methods. \textbf{Method 1.} Suppose $f(a)=f(b)$ holds, then comparing $P(x,y,a)$ and $P(x,y,b)$ we get $$2f((x-y)a)=2f((x-y)b)\Rightarrow f(ta)=f(tb) \enspace \forall t \in \mathbb {R}$$Let $u=\frac{a}{b}$, we obtain that $f(x)=f(ux)$ for all real $x$. We will now prove that $u=\pm 1$. Consider $P(ux, y, 1)$ and $P(x, uy, 1)$, then their difference is simply the equation $f(ux-y)=f(x-uy)$. Till this end, simply note that if $u^2\neq 1$, then for every pair of real numbers $p, q\in \mathbb{R}$, there exist $x, y$ such that $p=ux-y, q=uy-x$ by taking $x=\frac{up+q}{u^2-1}, y=\frac{uq+p}{u^2-1}$. Hence $f(p)=f(q)$ for all $p, q\in \mathbb{R}$ implying $f$ is constant. Plugging into the original equation we see that only the zero function works. Otherwise $u^2=1\Rightarrow a=\pm b$. $\square$ \textbf{Method 2.} We prove that if there exist $a,b>0$ with $f(a)=f(b)$, then $f$ is the zero function. Then by even property of $f$ we can deduce the same conclusion as in Method $1$. As above we prove that $f(a)=f(b)\Rightarrow f(ta)=f(tb) \Rightarrow f(1)=f(\frac{a}{b})$. This time we direct our attention on the set $$U=\{t|t=\frac{a}{b}, f(a)=f(b)\}$$Then it is clear that $t\in U \Rightarrow t^k\in U \enspace \forall k\in \mathbb{Z}$. Let $u=\frac{a}{b}$, then it suffices to prove that for every $t\in [1,u), t\in U$, or for every $t\in (\frac{1}{u},1], t\in U$. This is because then every number in $[1,u^n), (\frac{1}{u^n},1]$ will be in $U$ for all positive integers $n$, hence every positive reals will be in $U$. Now, suppose that there exist $z$ with $f(z)\neq 0$. Compare $P(ta,ta,z)$ with $P(tb,tb,z)$ we obtain that $$f(2t^2a^2+f(z))=f(2t^2b^2+f(z))\Rightarrow \frac{2t^2a^2+f(z)}{2t^2b^2+f(z)}\in U$$We split into two cases: - If $f(z)< 0$, then let $t$ vary from $[0, \sqrt{\frac{-f(z)}{2b^2}})$, then by continuity of the function $g(t)=\frac{2t^2a^2+f(z)}{2t^2b^2+f(z)}$ in $t$, when $t\in [0, \sqrt{\frac{-f(z)}{2b^2}})$, range of $g(t)$ must includes $[g(0), g(\sqrt{\frac{-f(z)}{2b^2}}))$, which is $[1, +\infty)$ or $(-\infty, 1]$. In either case, $[1,u)$ or $(\frac{1}{u},1]$ is in the range of $g$, which is in $U$. - If $f(z)>0$, then $\frac{2t^2a^2+f(z)}{2t^2b^2+f(z)}=(\frac{a^2}{b^2})(1-\frac{(1-\frac{b^2}{a^2})f(z)}{2t^2b^2+f(z)})$, so by varying $t$ from $[0,+\infty)$, the fraction will take every value in $[1,\frac{a^2}{b^2})$. So $[1, \frac{a}{b})=[1,u)$ is in $U$. So every positive reals are in $U$, which means $f(t)=f(1)$ for all $t>0$. By even property of $f$, $f(t)=f(1) \enspace \forall t\neq 0$. Now if $f(t)\neq 0$ for $t\neq 0$, then $P(t,0,t)\Rightarrow f(2f(t))=2f(f(t))+2f(t^2) \Rightarrow f(t)=4f(t)\Rightarrow f(t)=0$, false. So $f(t)=0$ for all $t \in \mathbb{R}$. $\square$ \textbf{Method 3.} Finally we present a short method to this claim. We proceed as in Method 1 that $f(a)=f(b)\Rightarrow f(a^2)=f(ab)=f(b^2)$. Then comparing $P(ab, ab, z)$ and $P(a^2, b^2, z)$ we get $$P(ab,ab,z)\Rightarrow f(f(ab)+f(ab)+f(z))=f(0)+f(2a^2b^2+f(z))+2f(0)$$$$P(a^2,b^2,1)\Rightarrow f(f(a^2)+f(b^2)+f(z))=f(f(a^2)-f(b^2))+f(2a^2b^2+f(z))+2f((a^2-b^2)z) $$Since $a^2\neq b^2$, both equation simplifies to $$f((a^2-b^2)z)=f(0) \Rightarrow f(t)=0 \enspace \forall t \in \mathbb{R}$$So $f(a)=f(b)\Rightarrow a=\pm b$. $\square$ Now we finish the problem. Suppose $f$ is not the zero function, consider $P(x,x,z)$ we obtain $f(2f(x)+f(z))=f(2x^2+f(z)) \Rightarrow f(x)=x^2$ or $f(z)=x^2-f(x)$ for all $z$. The later implies $f$ is constant which cannot be true since $f$ is not the zero function. So $f(x)=x^2$ for all $x\in \mathbb{R}$. Check: $f(x)=0$ gives $0=0+0+2(0)$, true. $f(x)=x^2$ gives $RHS=(x^2-y^2)^2+(2xy-z^2)^2+2(xz-yz)^2=x^4+y^4+z^4+2x^2y^2+2y^2z^2+2z^2x^2=(x^2+y^2+z^2)^2=LHS$, true. So both solutions are indeed solutions. $\blacksquare$

Similar to the above We claim that the answers are $f \equiv 0, x^2.$ It's easy to verify that these indeed satisfy the given equation. Now, let $P(x, y, z)$ denote the assertion that $f(f(x) + f(y) + f(z)) = f(f(x) - f(y)) + f(2xy + f(z)) + 2f(xz-yz).$ By $P(x, x, z),$ we know that $f(2f(x) + f(z)) = f(2x^2 + f(z)) + 3f(0).$ By $P(x, y, z) - P(y, x, z),$ we know that $f(f(x) - f(y)) + 2f((x-y)z) = f(f(y) - f(x)) + 2f((y-x)z).$ Comparing this with $P(x, y, -z) - P(y, x, -z)$ and letting $x \neq y, z$ vary, we know that $f$ is even. By $P(x, y, z) - P(x, -y, z),$ we know that $f(-2xy + f(z)) + 2f(z(x+y)) = f(2xy + f(z)) + 2f(z(x-y)).$ Let the assertion above be $Q(x, y, z).$ Suppose that there exists some $z_1, z_2$ such that $f(z_1) = f(z_2),$ with $z_1^2 \neq z_2^2.$ Notice that whenever $a^2 - b^2 = c^2 - d^2,$ we can get from $Q(a+b, a-b, z_1), Q(c+d, c-d, z_2)$ that: $$f(2az_1) - f(2bz_1) = f(2cz_2) - f(2dz_2),$$ i.e., $f(w) - f(x) = f(y) - f(z)$ whenever $\frac{w^2 - x^2}{z_1^2} = \frac{y^2 - z^2}{z_2^2}.$ Consider $t_1$ and $t_2$ such that $t_1 ^2 \neq t_2^2,$ and WLOG assume that $t_1^2 > t_2^2, z_1 > z_2$ (up to here, we haven't made any such assumptions). Then, there clearly exists some $x \in \mathbb{R}$ such that $\frac{x^2 - t_1^2}{z_2^2} = \frac{x^2 - t_2^2}{z_1^2}.$ But then the above equation implies that $f(t_1) = f(t_2).$ Therefore, we've shown that whenever $t_1 ^2 \neq t_2^2$, $f(t_1) = f(t_2),$ clearly implying that $f$ is constant and hence is $0$. Else, let's assume that $f$ is nonconstant, and so hence $f(a) = f(b) \Rightarrow a = \pm b.$ Therefore, by $P(x, y, 0) - P(x, -y, z),$ we know that: $$f(-2xy + f(0)) = f(2xy + f(0)) \Rightarrow -2xy + f(0) = \pm (2xy + f(0)),$$ which clearly implies that $f(0) = 0.$ Now, $P(x, x, z)$ gives that $f(2f(x)+f(z)) = f(2x^2+f(z)),$ and so hence either $f(x) = x^2$ or $2f(x) + f(z) = -2x^2 - f(z) \Leftrightarrow f(x) = -x^2 - f(z),$ for all $x, z \in \mathbb{R}.$ By simply selecting $f(z)$ such that it is not $-x^2 - f(x),$ we quickly obtain that $f(x) = x^2,$ i.e. $f(x) \equiv x^2$ $\forall x \in \mathbb{R},$ as desired. $\square$

Easy for an APMO#5 I did it with two steps 1.f(0)=0 2.if f(a)=f(b) then a^2=b^2(f is not zero)

navi_09220114 wrote: $RHS=(x^2-y^2)^2+(2xy-z^2)^2+2(xz-yz)^2=x^4+y^4+z^4+2x^2y^2+2y^2z^2+2z^2x^2=(x^2+y^2+z^2)^2=LHS$, true. So both solutions are indeed solutions. $\blacksquare$ The results are different. \begin{align*} (X^2 - Y^2)^2 + (2 XY - Z^2)^2= 2 (XZ - YZ)^2 + (X^2 + Y^2 - Z^2)^2 \end{align*}

The main idea of this solution is very similar to that of Iran Finals 2016 A3 and RMM 2019/5 and APMO 2019/5 (see post #22). The answer is $f(x) \equiv 0$ and $f(x) \equiv x^2$. These clearly work. Let $P(x,y,z)$ be the given assertion. It is easy to note that $f(x) \equiv 0$ is the only constant solution, so henceforth assume $f$ is non-constant. Lemma: If $f(a) = f(b)$ for some $a,b \in \mathbb R$, then $a^2 = b^2$ ; i.e. $a = \pm b$. Proof: Suppose not. Wlog, $a \ne 0$. Comparing $P(x,y,a)$ and $P(x,y,b)$ gives $$f(\alpha x ) = f(x) ~ \forall ~ x \in \mathbb R$$where $\alpha = b/a$ (so $\alpha^2 \ne 1$). Then comparing $P(x,y,z)$ and $P(\alpha x, \alpha y, z)$ yields $$f(2\alpha^2 \cdot xy + f(z)) = f(2xy + f(z)) = f(2 \alpha^2 \cdot xy + \alpha^2 f(z))$$As $f(z) \ne 0$ for some $z \in \mathbb R$, so this implies there is a constant $c \ne 0$ such that $$f(x+c) = f(x) ~ \forall ~ x \in \mathbb R $$Then comparing $P(1,1,y)$ and $P(1+c,1,y)$ gives $$f(0) = f(cz) ~ \forall ~ c \in \mathbb R$$But that would imply $f$ is constant, which is a contradiction. This proves our claim. $\square$ Claim: $f$ is even. Proof: Comparing $P(x,x,z)$ and $P(-x,-x,z)$ gives $$f(2f(x) + f(z)) = f(2f(-x) + f(z)) ~ \forall ~ x,z \in \mathbb R$$Now fix $x$. As $f$ is non-constant, so we can choose $z$ s.t. $2f(x) + 2f(-z) + 2f(z) \ne 0$. Then using our lemma we get $$2f(x) + f(z) = 2f(-x) + f(z)$$which implies our claim. $\square$ Claim: $f(0) = 0$. Proof: Using $f$ is even and comparing $P(x,x,0)$ and $P(x,-x,0)$ gives $$f(2x^2 + f(0)) = f(-2x^2 + f(0)) ~ \forall ~ x \in \mathbb R$$Now just fix any non-zero $x$ and then our lemma forces $$(2x^2 + f(0)) + (-2x^2 + f(0)) = 0$$which implies our claim. $\square$ Now we are ready to finish the problem. $P(x,x,z)$ yields $$f(2f(x) + f(z)) = f(2x^2 + f(z)) ~ \forall ~ x,z \in \mathbb R$$Fix a $x$. As $f$ is non-constant, we can choose $z$ s.t. $2f(x) + 2x^2 + 2f(z) \ne 0$. Then our lemma gives $$2f(x) = 2x^2 ~ \implies f(x) = x^2$$This completes the proof of the problem. $\blacksquare$

I agree with @guptaamitu1. Clearly any constant solution is the identically vanishing one which works. So we reject any constant solution here forth. Denote the assertion by $P(x,y,z).$ We aim to show that $f(x)\equiv x^2,$ which clearly works. Claim 1: $f$ is even. Proof. Comparing $P(1,2,x)$ and $P(2,1,x)$ gives $f(x)-f(-x)=k.$ We can immediately get $k=0,$ hence the claim. $\blacksquare$ Claim 2: $f(a)=f(b)\implies f(ac)=f(bc).$ Proof. Follows from $P(c,0,a)$ and $P(c,0,b).$ $\blacksquare$ Claim 3: $f(a)=f(b)\implies a=\pm b.$ Proof. Let $f(a)=f(b)$ with $|a|\neq |b|.$ Consider combining $P(ax,by,1)$ with $P(bx,ay,1)$ we have $f(ax-by)=f(bx-ay)=k,$ contradiction. $\blacksquare$ Claim 4: $f(0)=0.$ Proof. We compare $P(a,b,0)$ and $P(a,-b,0)$ so that $f(0)+2ab\in \{f(0)-2ab, 2ab-f(0)\}.$ Note that the former forces $ab=0,$ absurd. $\blacksquare$ Claim 5: $f(x)=x^2.$ Proof. Assume $\exists x: f(x)\neq x^2.$ Then $P(x,x,y)$ implies $2f(x)+f(y)\in \{f(y)+2x^2,-(f(y)+2x^2)\}.$ Thus $f(y)=-(f(x)+x^2)=k,$ absurd. $\blacksquare$ We are done.

I solved this on airplane (with paper) Given how sleepy I am currently I probably made typos The only solutions are $\boxed{f\equiv 0}$ and $\boxed{f(x)=x^2}$. The solution $f\equiv 0$ clearly works. Claim: The solution $f(x)=x^2$ works. Proof: If $f(x)=x^2$, then the LHS becomes $f(x^2+y^2+z^2)$, which is equal to \[x^4+y^4+z^4+2x^2y^2+2y^2z^2+2x^2z^2\] The RHS is \begin{align*} f(x^2-y^2)+f(2xy+z^2)+2x^2z^2-4xyz^2+2y^2z^2 \\ =x^4-2x^2y^2+y^4 + 4x^2y^2+4xyz^2 + z^4 + 2x^2z^2 - 4xyz^2 + 2y^2z^2 \\ =x^4+y^4+z^4+2x^2y^2+2y^2z^2+2x^2z^2, \end{align*}the same as the LHS, as desired. $\square$ It remains to show that those are the only solutions. Clearly the only constant solution is $f\equiv 0$, so henceforth assume $f$ is nonconstant. Claim: $f$ is even ($f(x)=f(-x)\forall x\in \mathbb{R}$). Proof: $P(x,y,0): f(f(x)+f(y)+f(0))=f(f(x)-f(y))+f(2xy+f(0))+2f(0)$. $P(y,x,0): f(f(x)+f(y)+f(0))=f(f(y)-f(x))+f(2xy+f(0))+2f(0)$. So $f(f(x)-f(y))=f(f(y)-f(x))$. $P(x,0,1): f(f(x)+f(0)+f(1))=f(f(x)-f(0))+f(f(1))+2f(x)$. $P(0,x,1): f(f(x)+f(0)+f(1)) = f(f(0)-f(x))+f(f(1))+2f(-x)$. Since $f(f(x)-f(0))=f(f(0)-f(x))$, we have $2f(x)=2f(-x)\implies f(x)=f(-x)$. $\square$ Claim: If $f(x)=f(0)$, then $x=0$. Proof: Suppose FTSOC there existed $k\ne 0$ with $f(k)=f(0)$. $P(x,0,0): f(f(x)+2f(0))=f(f(x)-f(0))+f(f(0))+2f(0)$. $P(x,0,k): f(f(x)+2f(0))+f(f(x)-f(0))+f(f(0))+2f(xk)$. Thus, $2f(0)=2f(xk)\implies f(0)=f(xk)$. Since $k\ne 0$, $xk$ can take on any real value, so $f$ is constant, contradiction. $\square$ Define $f$ to be partially injective if the following condition holds for all reals for all reals $a,b$: If $f(a)=f(b)$, then either $a=b$ or $a=-b$. Claim: $f$ is partially injective. Proof: Suppose FTSOC there exists $a,b$ where $f(a)=f(b), a\ne b, a\ne -b$ all hold true. By our earlier claim, $f(x)=f(0)$ iff $x=0$, so we can assume $a$ and $b$ are nonzero. Since $f$ is even, $f(|a|)=f(|b|)$, so we can assume $a,b>0$. $P(x,0,a): f(f(x)+f(0)+f(a))=f(f(x)-f(0))+f(f(a))+2f(ax)$. $P(x,0,b): f(f(x)+f(0)+f(a))=f(f(x)-f(0))+f(f(a))+2f(bx)$. Thus, $2f(ax)=2f(bx)\implies f(ax)=f(bx)$. Setting $x\to \frac{x}{a}$ since $a>0$ gives $f(x)=f\left(x\cdot \frac{b}{a}\right)$ for any $x\in \mathbb{R}$. $P\left(x,\frac{1}{2},z\right): f\left(f(x)+f\left(\frac{1}{2}\right)+f(z)\right) = f\left(f(x)-f\left(\frac{1}{2}\right)\right)+f(x+f(z))+2f\left(xz-\frac{z}{2}\right)$. $P\left(\frac{xb}{a}, \frac{b}{2a}, z\right); f\left(f\left(\frac{xb}{a}\right) + f\left(\frac{b}{2a}\right) + f(z) \right) = f\left(f\left(\frac{xb}{a}\right) - f\left(\frac{b}{2a}\right) \right) + f\left(x\cdot \frac{b^2}{a^2} + f(z)\right) + 2f\left(\frac{xbz}{a}-\frac{zb}{za}\right)$ Note that $f\left(\frac{1}{2}\right)=f\left(\frac{b}{2a}\right)$ and $f\left(xz-\frac{z}{2}\right) = f\left(\frac{xbz}{a}- \frac{zb}{za} \right)$. Using this, comparing with the $P\left(x,\frac{1}{2}, z\right)$ gives \[f\left(x\cdot \frac{b^2}{a^2} + f(z)\right) = f(x+f(z))\] Setting $z$ such that $f(z)\ne 0$ (since $f$ is nonconstant), and $x=-f(z)$ gives \[f\left(-f(z)\cdot \frac{b^2}{a^2} + f(z)\right)=f(0)\implies -f(z)\cdot \frac{b^2}{a^2} + f(z)=0\implies -\frac{b^2}{a^2} + 1 =0\implies \frac{b^2}{a^2} =1 \] However, this gives $a^2=b^2\implies a=b$ or $a=-b$, contradiction. Thus, $f$ is partially injective. $\square$ Claim: $f(0)=0$. Proof: $P\left(x,\frac{1}{2},0\right): f\left(f(x)+f\left(\frac{1}{2}\right) + f(0)\right) = f\left( f(x) - f\left(\frac{1}{2}\right)\right)+ f(x+f(0))$ $P\left(-x,\frac{1}{2},0\right): f\left(f(x)+f\left(\frac{1}{2}\right) + f(0)\right) = f\left( f(x) - f\left(\frac{1}{2}\right)\right)+ f(-x+f(0))$. Thus, $f(x+f(0))=f(-x+f(0))$ for any real $x$. So $f(f(0)+1)=f(f(0)-1)$. This implies $f(0)+1=f(0)-1$ or $f(0)+1=-(f(0)-1)=-f(0)+1$. The former implies $2=0$, not true. Thus, the latter must hold true. This implies $f(0)=0$. $\square$ $P(x,x,0): f(2f(x))=f(2x^2)$. So either $2f(x)=2x^2$ or $2f(x)=-2x^2$. Thus, for each $x$ either $f(x)=x^2$ or $f(x)=-x^2$. Claim: $f(x)=x^2\forall x\in \mathbb{R}$. Proof: Suppose FTSOC there existed $c\ne 0$ with $f(c)=-c^2$ $P(c,c,c): f(3f(c))=f(2c^2+f(c))\implies f(-3c^2)=f(c^2)$. Thus, $c^2=3c^2$ or $-3c^2$, both imply $c=0$, contradiction. So $f(x)=x^2\forall x\ne 0$ and $f(0)=0=0^2$. $\square$

Very similiar to 2019 APMO 5... the key ideas of showing $f(ax) = f(bx)$ and then demonstrating injectivity over $\mathbb{R}^{+}$ are nearly identical...

If $f(a) = f(b)$, where $a \neq b$ and $a \neq -b$, then $P(x, y, a)$ and $P(x, y, b)$ gives $f(at) = f(bt)$, so $f(x) = f(rx)$ for some $r$, with $|r| > 1$. $P(x, y, z)$ being compared to $P(rx, ry, z)$ gives $f(2r^2xy+f(z)) = f(2xy+\frac{f(z)}{r^2}) = f(2xy+f(z))$. Fixing $x$ and $z$ and letting $y$ vary, we see $f$ is periodic. Finally, replacing $z$ by $z+T$, we see that $f$ is constant. $f \equiv 0$ is clearly the only constant solution, so assume henceforth $f$ is nonconstant Now, letting $z = 0$ and switching $x$ and $y$, $f(f(x)-f(y)) = f(f(y)-f(x))$, and now switching the order again, $f(a) = f(-a)$ for all $a$. Now, fixing $z = 0$ and replacing $y$ with $-y$, $f(2xy+f(0)) = f(-2xy+f(0))$m which implies $f(0) = 0$. Finally, $x = y$ implies $f(2f(x)+f(z)) = f(2x^2+f(z))$, so $f(x) = x^2$ or $f(x)+x^2+f(z) = 0$, which cannot be true by the non constant property of $f$.

Nice! The answer is $f \equiv 0$ and $f(x)=x^2$, which clearly work. I will now show these are the only ones. Clearly $f \equiv 0$ is the only constant solution, so suppose $f$ is nonconstant. Let $P(x,y,z)$ denote the assertion. By comparing $P(x,y,0)$ and $P(y,x,0)$ we find that $f(f(x)-f(y))=f(f(y)-f(x))$. Now by comparing $P(x,y,z)$ with $P(y,x,z)$ and using this fact, we obtain $f((x-y)z)=f((y-x)z)$. Since this holds for all $x,y,z \in \mathbb{R}$ we readily obtain $f$ even. Suppose that $f(a)=f(b)$ for some $a \neq b$. Then by comparing $P(x,0,a)$ and $P(x,0,b)$ we find that $f(ax)=f(bx)$ for all $x$. Thus let $S:=\{r \mid f(x)=f(rx)~\forall x \in \mathbb{R}\}$, so we have $a/b \in S$ (if $b=0$ then use $b/a \in S$ instead). On the other hand, for any $r \in S$ by comparing $f(x,y,z)$ with $f(rx,ry,z)$ implies that $$f(2xy+f(z))=f(2xyr^2+f(z)) \implies \frac{2xyr^2+f(z)}{2xy+f(z)}=r^2+\frac{(1-r^2)f(z)}{2xy+f(z)} \in S.$$Fix some $z$ such that $f(z) \neq 0$: if no such $z$ exists then $f \equiv 0$ and we're done. If $r^2 \neq 1$, then by varying $2xy$ across all of $\mathbb{R}$ we find that every real except for $r^2$ is in $S$. On the other hand, $S$ is closed under multiplication, so clearly $r^2 \in S$ as well. Thus we have $S \equiv \mathbb{R}$, so $f$ is constant. But since we supposed $f$ is nonconstant, no such $r$ exists, hence $f(a)=f(b) \implies |a|=|b|$ (i.e. $S=\{1,-1\}$). By comparing $P(x,x,0)$ and $P(x,-x,0)$ and using the fact that $f$ is even, we find that $f(-2x^2+f(0))+2f(0)=f(2f(x)+f(0))=f(2x^2+f(0))+2f(0)$. Thus $|-2x^2+f(0)|=|2x^2+f(0)|$ for all $f$, which clearly implies $f(0)=0$. Thus from $P(x,x,z)$ we have $f(2f(x)+f(z))=f(2x^2+f(z))$ for all $x,z \in \mathbb{R}$. If we fix some $x$, it follows that $|2f(x)+f(z)|=|2x^2+f(z)|$ for all $z$. If $2f(x)+f(z)=-2x^2-f(z) \implies f(z)=-x^2-f(x)$ for all $z$, then since $x$ is fixed, so is $-x^2-f(x)$, so $f$ is constant: contradiction. Hence some $z$ has $2f(x)+f(z)=2x^2+f(z) \implies f(x)=x^2$. Since $x$ is arbitrarily chosen, it follows that $f(x)=x^2$ for all $x$, which is the other advertised solution. $\blacksquare$

The solutions are $f\equiv 0$ and $f\equiv x^2$. We henceforth assume that $f$ is not constant. Comparing $P(x,y,0)$ and $P(y,x,0)$, we get $f(f(x) - f(y)) = f(f(y) - f(x))$. Now comparing $f(z(x-y)) = f(z(y-x))$, we get that $f(z(x-y)) = f(z(y-x)) \implies f$ is even. Therefore clearly $f(x) = f(-x)$ for all $x\neq 0$ and thus, $f$ is not injective. Therefore there exist $a,b$; $a\neq b$ such that $f(a) = f(b)$. Then comparing $P(x,y,a)$ and $P(x,y,b)$, we get that $f((x-y)a) = f((x-y)b) \implies f(x) = f(rx)$, $r\neq 1$ for all $x\in \mathbb R$. Now comparing $P(rx,r,z)$ and $P(x,1,z)$, we get $f(2r^2x + f(z)) = f(2x + f(z))$. Now pick $z_0$ such that $f(z_0) \neq 0$. Then for any $k\in \mathbb{R}$, $k\neq r^2$, we get, \[ k = \dfrac{2r^2x + f(z_0)}{2x + f(z_0)} \iff x = \dfrac{f(z_0) - kf(z_0)}{2k - 2r^2} \text{ if }r\neq -1. \] Then $f(x) = f(kx)$ for all $k\in \mathbb R$ and thus $f$ is constant, contradiction. Therefore we must have that $r=-1$, that is $a=-b$. If $f(0) = 0$, we can do the same by comparing $P(rx,y,0)$ and $P(x,y,0)$ to get that $f$ is constant, which works even if $r = -1$. Therefore $f(0) = 0$. Now $P(x,x,0) \implies f(2f(x)) = f(2x^2) \implies f(x) \in \left\{+x^2, -x^2\right\}$. If $f(b) = -b^2$, $P(b,b,z)$ gives a contradiction. Thus $f \equiv x^2$.

Let $P(x,y,z)$ denote the usual. Notice that the constant function $f(x)=0$ satisfies the equation, so henceforth assume $f$ is non-constant since the only constant solution is clearly $0$. $P(x,y,0): f(f(x)+f(y)+f(0))=f(f(x)-f(y))+f(2xy+f(0))+2f(0)$ and swapping $x,y$ yields $f(f(x)-f(y))=f(f(y)-f(x))$ Now swapping $x,y$ in the original equation then gives $f(az)=f(-az)$ for all reals $a,z$ where $a=x-y$, so that $f$ is even as desired. Comparing $P(x,y,0),P(-x,y,0)$, we obtain $f(-2xy+f(0))=f(2xy+f(0))$ Thus any 2 reals summing to $2f(0)$ return equal values. Suppose that $f(0)\neq 0$ Then $P(0,0,0)$ yields $f(3f(0))=f(0)+f(f(0))+2f(0)$. Noticing that $f(3f(0))=f(-f(0))=f(f(0))$, so that $f(0)=0$, a contradiction. Hence $f(0)=0$. Now we show $f$ is "injective". Suppose there exist $m,n$ with $|m|\neq |n|$ and $f(m)=f(n)$ Comparing $P(x,y,m), P(x,y,n)$, we obtain $f(am)=f(an)$ for all reals $a$. Comparing $P(xm,ym,z),P(xn,yn,z)$, we obtain $f(2xym^2+f(z))=f(2xyn^2+f(z))$ for all reals $x,y,z$ WLOG $m>n$ and writing $\frac{m^2}{n^2}=d$, we obtain $f(ad+f(z))=f(a+f(z))$ for all reals $a,z$. In fact, letting $a=-f(z)$ we obtain $f(f(z)(1-a))=0$ Lemma: $f(x)=0$ implies $x=0$ Proof: suppose that $f(c)=0$ Comparing $P(x,y,c),P(x,y,0)$, we obtain $f(0)=f(c(x-y))$ for all reals $x,y$. Since $f$ is non-constant, this must imply that $c=0$ as desired. Returning to the problem, we then obtain $f(z)(1-a)=0$, i.e. $1-a=0$, which is impossible since $|m|\neq |n|$. This implies that $f(m)=f(n) \Rightarrow |m|=|n|$ Now to finish, take $P(x,x,0) \Rightarrow f(2f(x))=f(2x^2)$ which implies that $f(x)=\pm x^2$ However, if $f(x)=-x^2$ then $P(x,x,x) \Rightarrow -9x^2=-x^2$, a contradiction for all $x \neq 0$ Hence we obtain $f(x)=x^2$ which is indeed a solution. Thus, our solutions are the constant function $f(x)=0$ and $f(x)=x^2$

The answer is $f(x) = \boxed{0,x^2}$, which works. Let the given assertion be $P(x,y,z)$. If $f$ is constant, it must equal $0$, so assume $f$ is nonconstant henceforth. Claim: $f$ is injective Proof: Suppose that there exist distinct nonzero numbers such that $f(a) =f(b)$. Comparing $P(x,0,a)$ and $P(x,0,b)$ yields $f(ax) = f(bx)$. Therefore, we have $f(x) = f(rx)$ where $r = \tfrac{a}{b}$. Then, comparing $(x,y,0)$ and $P(y,x,0)$ yields \[f(f(x)-f(y)) = f(f(y)-f(x)),\] which implies that $f$ is even. Comparing $P(rx,y,z)$ and $P(x,ry,z)$ gives \[f(z(rx-y)) = f(z(x-ry)) = f(z(ry-x)).\] This implies that \[f(z) = f \left(\frac{rx-y}{ry-x} \cdot z \right),\] but $\tfrac{rx-y}{ry-x}$ can be any value for $x,y \in \mathbb{R}$. Thus, $f$ is constant, a contradiction. $\square$ Comparing $P(x,y,0)$ and $P(x,-y,0)$ yields \[f(2xy+f(0)) = f(2xy-f(0)).\] Injectivity then reveals that $f(0) = 0$. Finally, $P(x,x,z)$ yields \[f(2f(x)+f(z)) = f(2x^2+f(z))\]\[\implies 2f(x)+f(z) = \pm (2x^2+f(z)),\] leaving us with two cases. The positive case leads to our latter solution, so we will deal with the negative case. We know that $f(z) = f(x)-x^2,$ but fixing $x$ implies that $f$ is constant, a contradiction. Hence, the only solutions are those in our solution set.

Let $P(x, y, z)$ denote the assertion. \[P(x, y, z)\]\[ f(f(x)+f(y)+f(z)) = f(f(x)-f(y))+f(2xy+f(z))+2f(xz-yz). \]\[P(y, x, z)\]\[ f(f(x)+f(y)+f(z)) = f(f(y)-f(x))+f(2xy+f(z))+2f(yz-xz). \]Thus $f(f(x)-f(y))+2f(xz-yz)=f(f(y)-f(x))+2f(yz-xz)$, letting $z=0$ gives: $f(f(x)-f(y))=f(f(y)-f(x))$, so $f(x)=f(-x)$. \[P(x, y, 0)-P(-x, y, 0)\]\[f(2xy+f(0))=f(2xy-f(0))\]Thus either $f(0)=0$ or $f$ is periodic. First suppose $f(0)=0$. Now suppose that $f(a)=f(b)$ and $(\frac{a}{b})^2\neq 1$. Let $\frac{a}{b}=c$. \[P(x, y, cz)-P(x, y, z)\]\[f(c(zx-zy))=f(zx-zy)\]Thus $f(cx)=f(x)$. \[P(x, y, cz)-P(x, y, z)\]\[f(cxy+f(z))=f(xy+f(z))\]Thus if $f(z)\neq 0$, we can choose $x$ and $y$ such that $(\frac{cxy+f(z)}{xy+f(z)})^2$ attains all non negative real values. Thus as we have a value $a$ and $b$ such that $f(a)=f(b)$ where $\frac{a}{b}=k$ for all non negative reals $k$. Thus $f(k^2x)=f(x)$ for all non negative reals $k^2$ as $f(x)=f(-x)$ this covers means $f$ is constant. Thus if $f$ is constant $f(x)=0$ for all $x$. Thus if $f(a)=f(b)$, $\frac{a}{b}^2=1$. \[P(x, x, 0)\]\[f(2f(x))=f(2x^2)\]Thus $f(x)=\pm x^2$. Now suppose that we have infinetly many values such that $f(y)=y^2$, also suppose $f(x)=-x^2$. \[P(x, x, y)\]\[f(-2x^2+y^2)=f(2x^2+y^2)\]This is clearly a contradiction as we can choose $y\neq0$. Thus $f(x)=x^2$ for all $x$ or $f(x)=-x^2$. However $-x^2$ clearly does not work. Now I will deal with when $f$ is periodic. Let the period be $c$. \[P(c, y, 1)-P(0, y, 1)\]\[f(yc+f(1))=f(f(1))\]Thus as $yc$ can take on any value $f$ must be constant so $f(0)=0$ for all $x$ or $f(x)=x^2$ for all $x$.

Take $(x, y, z), (y, x, z)$ and then compare the values at $z = 1, z = -1$ to get $f$ is even. Let $S = \{r\mid f(rx) = f(x)\}.$ If $S\neq\{1, -1\},$ then let us assume $f(a) = f(b).$ Take $(1, 0, bx)$ and then $(1, 0, ax)$ to get that $f(ax) = f(bx).$ Now, take $(x, y, z)$ and compare to $(rx, ry, r)$ to get $f(2r^2xy + f(z)) = f(2xy + f(z)).$ If $r = 0,$ then $f(f(z)) = f(2xy + f(z)),$ telling us $f$ is constant, and thus 0. Otherwise, $f(2r^2xy + f(z)) = f(2xy + f(z)) = = f(2xy + \frac1{r^2}f(z)).$ If $f$ wasn't 0, then we have that $f(x) = f(x + c)$ for some constant $c.$ Now, take $(x + c, y, z)$ and compare to $(x, y, z)$ for some $z\in\mathbb Z$ to get $f(2xy + 2yc + f(z)) = f(2xy + f(z)).$ By changing values of $y$ and fixing $xy,$ we can get that $f$ is constant, and thus 0. Now, we can assume $f$ is injective over $\mathbb R^+.$ Take $(x, x, z)$ to get $f(2f(x) + f(z)) = f(2x^2 +f(z)).$ Hence, either $2f(x) + f(z) = 2x^2 + f(z)$ or $2f(x) + f(z) = -2x^2 - f(z).$ The second one implies that $f(z) = 2f(x) + 2x^2$, or $f$ is constant(impossible) while the first implies $f(x) = x^2.$ Thus, $f\equiv x^2.$