Here is my quite lengthy solution:

I try to shorten it a little bit: 1st part, we show $f(0)=0$, $f$ is bijective and $f(x)=-f(-x)$:

3rd part, conclusion

I don't understand how $ x^{2}=f(xf(z))-f((z-x)f(x)) $ and $f(x)=f(z)$ implies $z-x=0$ or $x=0$. Could you explain?

tenniskidperson3 wrote: I don't understand how $ x^{2}=f(xf(z))-f((z-x)f(x)) $ and $f(x)=f(z)$ implies $z-x=0$ or $x=0$. Could you explain? Since $f(z)=f(x)$, then $f(xf(z))=f(xf(x))=x^2$, or $f((z-x)f(x))=0$, ie by previous results $(z-x)f(x)=0$, yielding either $z=x$ or $f(x)=0$, this latter one being equivalent again to $x=0$. Hope this clears it out...

daniel73 wrote: tenniskidperson3 wrote: I don't understand how $ x^{2}=f(xf(z))-f((z-x)f(x)) $ and $f(x)=f(z)$ implies $z-x=0$ or $x=0$. Could you explain? Since $f(z)=f(x)$, then $f(xf(z))=f(xf(x))=x^2$, or $f((z-x)f(x))=0$, ie by previous results $(z-x)f(x)=0$, yielding either $z=x$ or $f(x)=0$, this latter one being equivalent again to $x=0$. Hope this clears it out... What if $f((z-x)f(x)=x^2 $? It also imply $z=x$ or $x=0$?

lefao wrote: daniel73 wrote: tenniskidperson3 wrote: I don't understand how $ x^{2}=f(xf(z))-f((z-x)f(x)) $ and $f(x)=f(z)$ implies $z-x=0$ or $x=0$. Could you explain? Since $f(z)=f(x)$, then $f(xf(z))=f(xf(x))=x^2$, or $f((z-x)f(x))=0$, ie by previous results $(z-x)f(x)=0$, yielding either $z=x$ or $f(x)=0$, this latter one being equivalent again to $x=0$. Hope this clears it out... What if $f((z-x)f(x)=x^2 $? It also imply $z=x$ or $x=0$? You have a bracket missing; if you mean $f((z-x)f(x))=x^2$, then we know by previous results that $x^2=f(xf(x))$, and that $f$ is injective, or you may deduce that $f(z-x)f(x)=xf(x)$, yielding either $f(x)=0$ (which again results in $x=0$), or $f(z-x)=x$. Since we have chosen $x\neq0$, you would obtain that $f((z-x)f(x))=x^2$ implies $f(z-x)=x$. However, I do not know under which circumstances you would obtain $f((z-x)f(x))=x^2$, and I do not see clearly why $f(z-x)=x$ would be useful, but if it is, feel free to use it.

Set $x = 0$. $f(0) = f(yf(0))$ for all $y$. If $f(0) \neq 0$, $yf(0)$ spans the set of reals, so $f$ must be constant, which is clearly impossible. Hence, $f(0) = 0$. Setting $y=0$ shows that $f(xf(x)) = x^2$ for all real $x$, and setting $x=-y$ shows that $f(-xf(x)) = -x^2$ for all real $x$. It follows from these that if $f(v) = 0$, then $v^2 = f(vf(v)) = f(0) = 0$, i.e., $v = 0$. We now claim that $f$ is injective. Suppose that $f(a) = f(a+b)$. We wish to show that $b=0$. From $f(af(a)) = a^2$, we have $a^2 = f(af(a)) = f(af(a+b)) = f(bf(a)) = a^2$, whence $f(bf(a)) = 0$. Hence, $b = 0$ or $f(a) = 0$. If $f(a) = 0$, then from the above $a = b = 0$, so $f$ is injective. From $f(xf(x)) = x^2$, we have $xf(x) f(xf(x)) = x^3 f(x)$, so $f(x^3 f(x)) = f((xf(x)) f(xf(x))) = x^2 f(x)^2$. When $x=1$, we have $1^2 = f(1 f(1)) = f(1^3 f(1)) = f(1)^2$. Similarly, $f(-1)^2 = 1$. By injectivity, either $f(1) = 1$ and $f(-1) = -1$, or $f(-1) = 1$ and $f(1) = -1$. Since $f(x)$ is a solution to the functional equation if and only if $f(-x)$ is (as we may replace each occurence of $x$ and $y$ in the original functional equation with $-x$ and $-y$), we may for now assume without loss of generality that $f(1) = 1$ and $f(-1) = -1$. Setting $y=x$ in the original functional equation gives $f(xf(2x)) = f(xf(x)) + x^2 = 2x^2$. Since $f(\sqrt{2} x f( \sqrt{2} x)) = 2x^2$ as well, by injectivity $x f(2x) = \sqrt{2} x f( \sqrt{2} x)$. Replacing $x$ with $\frac{\sqrt{2} x}{2}$ and rearranging gives $f(\sqrt{2} x) = \sqrt{2} f(x)$. Hence, $f(2x) = f( \sqrt{2} (\sqrt{2} x)) = \sqrt{2} f(\sqrt{2} x) = 2x$. Setting $x=1$ in the original functional equation gives $f(f(y+1)) - f(y) = 1$, and setting $x=2$ gives $f(f(y+2)) - f(y) = 2$. Subtracting these two equations gives $f(f(y+2)) - f(f(y+1)) = 1$ for all $y$, or $f(f(y+1)) = f(f(y)) + 1$. Substituting this into $f(f(y+1)) - f(y) = 1$ yields $f(f(y)) = f(y)$. By injectivity, $f(y) = y$ for all real $y$. Recall that this is only the solution when $f(1) = 1$; if $f(1) = -1$, $f(-x)$ is the solution to this functional equation, so $f(x) = x$ and $f(x) = -x$ are the only solutions.

Information by Kazuaki Kobayashi, MOF (Mathematic Olympiad Foudation) of Japan about the background for this problem.

bigbang195 wrote: Find all functions $f$ from the set of real numbers into the set of real numbers which satisfy for all $x$, $y$ the identity \[ f\left(xf(x+y)\right) = f\left(yf(x)\right) +x^2\] Proposed by Japan Here is a long and rather ugly solution. Let $P(x,y)$ be the assertion $f(xf(x+y))=f(yf(x))+x^2$ 1) $f(x)=0$ $\iff$ $x=0$ ============== If $f(0)\ne 0$, then $P(0,\frac x{f(0)})$ $\implies$ $f(x)=f(0)$ constant, which is not a solution. So $f(0)=0$ If $f(u)=0$, then $P(u,-u)$ $\implies$ $u^2=0$ Q.E.D. 2) $f(x)$ is a bijection ============= $P(x,0)$ $\implies$ $f(xf(x))=x^2$ and so $\mathbb R^+\cup\{0\}\subseteq f(\mathbb R)$ $P(x,-x)$ $\implies$ $f(-xf(x))=-x^2$ and so $\mathbb R^-\cup\{0\}\subseteq f(\mathbb R)$ And so $f(x)$ is a surjection Let then $a<0$ and $u$ such that $f(u)=a$ : $P(\sqrt{-a},\frac u{f(\sqrt{-a})})$ $\implies$ $f(\sqrt{-a}f(\sqrt{-a}+\frac u{f(\sqrt{-a})}))=0$ So (using 1. above) : $f(\sqrt{-a}+\frac u{f(\sqrt{-a})})=0$ So (using again 1) above) : $\sqrt{-a}+\frac u{f(\sqrt{-a})}=0$ and $u=-\sqrt{-a}f(\sqrt{-a})$ And so a unique $u$ for any $a<0$ Let then $a>0$ and $u\ne 0$ such that $f(u)=a$ and $v$ such that $f(v)=\frac u{\sqrt a}$ : $P(\sqrt a,v-\sqrt a)$ $\implies$ $f((v-\sqrt a)f(\sqrt a))=0$ So (using 1. above) : $(v-\sqrt a)f(\sqrt a)=0$ and so $v=\sqrt a$ and $u=\sqrt af(\sqrt a)$ And so a unique $u$ for any $a>0$ And so $f(x)$ is an injection. Q.E.D. 3) $f(x)$ is an odd function ================= Let $x\ne 0$ $P(x,0)$ $\implies$ $f(xf(x))=x^2$ $P(-x,0)$ $\implies$ $f(-xf(-x))=x^2$ And so $f(xf(x))=f(-xf(-x))$ and so, since injective, $xf(x)=-xf(-x)$ and $f(-x)=-f(x)$ (true also if $x=0$) Q.E.D. 4) $f(2x)=2f(x)$ ========= Let $x\ne 0$ : $P(x,-2x)$ $\implies$ $-f(xf(x))=-f(2xf(x))+x^2$ $P(-x,-x)$ $\implies$ $f(xf(2x))=f(xf(x))+x^2$ Subtracting, we get : $f(xf(2x))=f(2xf(x))$ and, since injective : $xf(2x)=2xf(x)$ and so $f(2x)=2f(x)$ (true also if $x=0$) Q.E.D. 5) $f(x+y)=f(x)+f(y)$ $\forall x,y$ ===================== $P(x,y)$ $\implies$ $f(xf(x+y))=f(yf(x))+x^2$ $P(x+y,-x)$ $\implies$ $f((x+y)f(y))=-f(xf(x+y))+(x+y)^2$ Subtracting, we get a new assertion $Q(x,y)$ : $f((x+y)f(y))+f(yf(x))=y^2+2xy$ $Q(x-y,y)$ $\implies$ $f(xf(y))+f(yf(x-y))=2xy-y^2$ $Q(x-y,2y)$ $\implies$ $f((x+y)f(y))+f(yf(x-y))=2xy$ Subtracting, we get $f(xf(y)+yf(y))-f(xf(y))=y^2$ And so $f(xf(y)+yf(y))=f(xf(y))+f(yf(y))$ (remember that $P(y,0)$ $\implies$ $f(yf(y))=y^2$) And so $f(x+yf(y))=f(x)+f(yf(y))$ $\forall x,y$ And so $f(x+y)=f(x)+f(y)$ $\forall x\in\mathbb R,\forall y\in f^{[-1]}(\mathbb R^+\cup\{0\})$ But then $f(-x+y)=-f(x)+f(y)$ and so $f(x-y)=f(x)+f(-y)$ And so $f(x+y)=f(x)+f(y)$ $\forall x\in\mathbb R,\forall y\in f^{[-1]}(\mathbb R^-\cup\{0\})$ Q.E.D. 6) either $f(x)=x$ $\forall x$, either $f(x)=-x$ $\forall x$ ================================ Using $f(x+y)=f(x)+f(y)$, $P(x,y)$ becomes $f(xf(x))+f(xf(y))=f(yf(x))+x^2$ and so $f(xf(y))=f(yf(x))$ $P(x+y,0)$ $\implies$ $f(xf(x))+f(xf(y))+f(yf(x))+f(yf(y))=(x+y)^2$ and so $f(xf(y))=xy$ Setting $y=f(1)$ and using $f(yf(y))=y^2$, we get $f(x)=xf(1)$ Plugging this in original equation, we get $f(1)^2=1$ Hence the two solutions : either $f(1)=1$ and so $\boxed{f(x)=x}$ $\forall x$ either $f(1)=-1$ and so $\boxed{f(x)=-x}$ $\forall x$

Another proof of injectivity: $f(x)=f(y)$ implies $x^2=f(xf(x))=f(xf(y))=f((y-x)f(x))+x^2$, which gives $f((y-x)f(x))=0$. So $y=x$ or $x=0$, both of which implies $x=y$. Another way to continue from (3): $f(xf(y))=f((y-x)f(x))+x^2=-f((x-y)f(x))+x^2=-f(yf(x-y))-(x-y)^2+x^2=f(yf(y-x))-(x-y)^2+x^2=f(-xf(y))+y^2-(x-y)^2+x^2=-f(xf(y))+2xy$ so $f(xf(y))=xy$, and similarly we get $f(yf(x))=xy$. Hence $f(xf(y))=f(yf(x))$, which gives $xf(y)=yf(x)$. So $f(x)=cx$, and we easily get $c=\pm1$.

Johan Gunardi wrote: Another proof of injectivity: $f(x)=f(y)$ implies $x^2=f(xf(x))=f(xf(y))=f((y-x)f(x))+x^2$, which gives $f((y-x)f(x))=0$. So $y=x$ or $x=0$, both of which implies $x=y$. Another way to continue from (3): $f(xf(y))=f((y-x)f(x))+x^2=-f((x-y)f(x))+x^2=-f(yf(x-y))-(x-y)^2+x^2=f(yf(y-x))-(x-y)^2+x^2=f(-xf(y))+y^2-(x-y)^2+x^2=-f(xf(y))+2xy$ so $f(xf(y))=xy$, and similarly we get $f(yf(x))=xy$. Hence $f(xf(y))=f(yf(x))$, which gives $xf(y)=yf(x)$. So $f(x)=cx$, and we easily get $c=\pm1$. Yesssss, quite OK for both parts and quite nice ! Much quicker and simpler than my ugly steps. Congrats !

No need for injectivity ! $P(0,y) \implies f(0)=f(yf(0))$, and if f(0) is not zero, then f is constant and we get a contradiction. Hence f(0)=0. $P(x,0) \implies f(xf(x))=x^2$ and $P(x,-x) \implies f(-xf(x))=-x^2$ and in particular, f is onto. 1) $f(x)=0 \iff x=0$ Let $c\in \mathbb{R}$ such that f(c)=0. $P(c,0) \implies f(cf(c))=c^2 \implies 0=c^2 \implies c=0$ 2) $f(d)=1 \implies d=\pm 1$ Let $d\in \mathbb{R}$ such that f(d)=1. $P(d,0) \implies f(df(d))=d^2 \implies 1=d^2 \implies d=\pm 1$ 3) $f(x)=y^2 \implies x=yf(y)$ Let x,y $\neq 0$ and define $s=\frac{x}{f(y)}$. $P(s,y-s) \implies f(x)=f((y-s)f(s))+s^2$ If f(x)=s², then f((y-s)f(s))=0, hence y-s=0 or f(s)=0, y=s or s=0, y=s, f(y)=f(s), x=sf(s). As f is onto, s take all the real values when y changes in $\mathbb{R}$, and hence the desired. 4) $f(2x)=2f(x)$ $P(x,x) \implies f(xf(2x))=f(xf(x))+x^2=2x^2=(\sqrt{2}x)^2 \\ \implies xf(2x)=\sqrt{2}xf(\sqrt{2}x) \\ \implies f(2x)=2f(x)$ Now from f(f(1))=1 and 2), we see that f(1)=1 or f(1)=-1. 1) First case : f(1)=1 $f(2x)=2f(x) \implies f(2)=2$ $P(1,y) \implies f(f(y+1))=f(y)+1$ $P(2,y) \implies f(2f(y+2))=f(2y)+4 \\ \implies f(f(y+2))=f(y)+2 \\ \implies f(y+1)+1=f(y)+2 \\ \implies f(y+1)=f(y)+1$ Now, $f(y)+1=f(f(y+1))=f(f(y)+1)=f(f(y))+1 \implies f(y)=f(f(y))$ and since f is onto we get f(x)=x for all real x. 2) Second case : f(1)=-1 $f(2x)=2f(x) \implies f(2)=-2$ $P(1,y) \implies f(f(y+1))=f(-y)+1$ $P(2,y) \implies f(2f(y+2))=f(-2y)+4 \\ \implies f(f(y+2))=f(-y)+2 \\ \implies f(-y-1)+1=f(-y)+2 \\ \implies f(-y-1)=f(-y)+1 \\ \implies f(y-1)=f(y)+1 \implies f(y+1)=f(y)-1$ Now, $f(-y)+1=f(f(y+1))=f(f(y)-1)=f(f(y))+1 \implies f(f(y))=f(-y)$. $f(x^2)=f(f(xf(x))=f(-xf(x))=-x^2$ and$ f(-x^2)=f(f(-xf(x))=f(xf(x))=x^2$ so f(x)=-x for all real x. Hence the two solutions : f(x)=x and f(x)=-x.

Sorry for not translating. Sea la ecuación original $(1)$. Con $x=0$ en $(1)$ obtenemos que $f(0)=0$ ya que $f(yf(0))=f(0)$ y $f$ no puede ser constante. Con $y=0$ en $(1)$ obtenemos $f(xf(x))=x^2$, ecuación a la que llamaré $(2)$, y llamemos $g(x)=xf(x)$. Con $y=-x$ en $(1)$ obtenemos que $f(-xf(x))=-x^2$, ecuación a la que llamaré $(3)$. Con ésto, tenemos ya que $f$ es suryectiva. Supongamos que existe $a \neq 0$ tal que $f(a)=0$. Entonces en $(1)$ con $a=x$ tendremos que $a^2=f(af(a+y))$ para toda $y$. Tomemos $e_0$ un número distinto de $a^2$. Por suryectividad existe $e_1$ tal que $f(e_1)=e_0$. Luego, existe $e_2$ tal que $f(e_2)=e_1/a$. Con $y=e_2-a$ tendremos que $e_0=a^2$, contradicción. Entonces $f(a)=0$ sii $a=0$. Ahora supongamos $f(a)=f(b)$, y sea $k=f(a)$. En $(1)$ con $x=a$, $y=b-a$ tendremos, por $(2)$, que $a^2=f(af(a))=f(af(b))=f(xf(x+y))=f(yf(x))+a^2$ y entonces $f(yf(x))=0$. Por tanto, $yf(x)=0$. Si $f(x)=0$ entonces $f(a)=f(b)=0$ entonces $a=b=0$. De lo contrario, $y=0$ y así $a=b$. Obtenemos que $f$ es inyectiva. Por tanto, $f$ es biyectiva. Por $(2)$, $f(g(x))=x^2$ y $f(-g(x))=-x^2$. Con ésto es fácil ver que $g$ es suryectiva (dada la inyectividad de $f$). Sea $a$ un real cualquiera. Si $a=0$ tengo $f(0)=-f(-0)$. Sea $b$ tal que $g(b)=a$. Entonces vemos que $f(a)=-f(-a)$. De aquí, $f$ es impar. Ahora mostraré que $f(nxf(-x))=nx^2$ para todo $n \ge 0$ entero, usando inducción sobre $n$. Para $n=0$ es obvio. Lo demostraré para $n$ usando que es cierto para $n-1$. Tenemos por imparidad y por $(2)$ que $f((n-1)g(x))=(n-1)f(g(x))$ para todo $x$. De aquí vemos que $f((n-1)x)=(n-1)f(x)$ dado que $g$ es suryectiva. Ahora, en $(1)$ sea $y=na$ y $x=-a$ para cualquier $a$. Tendremos que $-a^2(n-1)=f((n-1)af(-a))=f(-af((n-1)a))=f(xf(x+y))=a^2+f(nxf(-x))$ Y de aquí, $f(nxf(-x))=-nx^2$ para toda $a$. La inducción está completa. Entonces, usando el argumento de arriba, pues $g$ es suryectiva, encontramos que $f(na)=nf(a)$ para todo $n \ge 0$ entero. Por imparidad vemos fácilmente que $f(ra)=rf(a)$ para todo $r$ racional, para todo $a$ real. A ésta ecuación le llamo $(4)$. En $(2)$ con $x=1$ y con $x=f(1)$ obtenemos que $f(f(1))=1$ y luego que $f(f(1))=f(1)^2$. Luego $f(1)=1$ ó $-1$. Dado que $f$ es impar, si defino $f_1(x)=-f(x)$, seguirá cumpliendo $(1)$, y entonces puedo suponer sin pérdida de generalidad que $f(1)=1$. De $(4)$, se da $f(r)=r$ para $r$ racional. De $(1)$ con $x=1$ observemos que $f(f(y+1))=f(y)+1$. A ésto le llamo $(5)$. Ahora en $(1)$ sea $x \neq 0$ un racional cualquiera y usemos $(5)$ y $(4)$. Resulta que $x(f(x+y-1)+1)=xf(f(x+y))=f(xf(x+y))=f(yf(x))+x^2$$=xf(y)+x^2$ y así $f(x+y-1)+1=f(y)+x$. Sea $r=x-1$, se da que $f(r+y)=f(y)+r$ para todo $r \neq -1$ racional y para todo $y$. A ésto le llamo $(6)$. Utilizemos $(5)$ y $(6)$. Con $r=1$ tenemos que $f(x+1)=1+f(x)$ y entonces $f(x)=1+f(x-1)$. Resulta que $f(f(x))=f(x-1)+1=f(x)$ para todo $x$. Por inyectividad tendremos $f(x)=x$ para toda $x$. Recordemos que supusimos sin pérdida de generalidad que $f(1)=1$, entonces en realidad $-f$ también cumple. Por ende, las únicas soluciones son: $f(x)=x$ $f(x)=-x$

Aiscrim wrote:

i think there's mistake,shouldn't there be :$f(f(x)f(\alpha))=f((\alpha-f(x))f(f(x)))+f(x)^2$ ? if i'm wrong,i apologize

You are right. My mistake, sorry

Let $P(x,y)$ be the assertion $f(x(f(x+y))=f(yf(x))+x^2$ Lemma 1 : $f(0)=0$ Proof : Let $f(0) \neq 0$ Consider $P(0,\frac{y}{f(0)})$ $$ \implies f(0)=f(y)$$That is $f \equiv f(0)$ But this satisfy the conditions of the given equation. Hence $\boxed{f(0)=0}$ Lemma 2 : $f(\alpha)=0 \implies \alpha =0$ Proof : Let if possible their exist some $\alpha \neq 0$ such that $f(\alpha)=0$ Consider $P(\alpha,0)$ $$\implies f(\alpha f( \alpha ))=f(0)+\alpha ^2$$$$\implies \alpha ^2 = 0$$A contradiction !!! Thus we must have only root of $f$ as $0$. Lemma 3 : $f$ is injective Proof : Let if possible there exist some $a \neq b$ and $f(a)=f(b)$ Then consider $P(b,a-b)$ $$\implies f(bf(a))=f((a-b)f(b))+b^2$$And now consider $P(b,0)$ $$\implies f(bf(b))=b^2$$ $$\implies f((a-b)f(b))=0$$By Lemma 2, $\implies (a-b)f(b)=0$ Thus we have either $a=b$ or $f(b)=0=f(a)$ which again implies $a=b$(by Lemma 2) A contradiction. Thus we get $f$ is injective. Lemma 4 : $f(xf(x))=x^2$ Proof : Just consider $P(x,0)$ Lemma 5 : $-f(s)=f(-s)$ Proof : For $x=0$ this is obvious. Now consider some $x \neq 0$ By Lemma 4, $$f(xf(x))=x^2=f(-xf(-x))$$Using injectivity, $$xf(x)=-xf(-x)$$$$\implies f(x)=-f(-x)$$Hence the lemma. Lemma 6 : $f(2x)=2f(x)$ Proof : Consider $P(x,x)$ $$\implies f(xf(2x))=f(xf(x))+x^2 = 2x^2 = f(x\sqrt{2}f(x\sqrt{2}))$$Injectivity implies $$xf(2x)=x\sqrt{2}f(x\sqrt{2})$$$$\implies f(2x)=\sqrt{2}f(x\sqrt{2})$$Replacing $x$ by $\frac{x}{\sqrt{2}}$, $$\implies f(x\sqrt{2}) = \sqrt{2}f(x)$$Putting this back, $$\boxed{f(2x)=2f(x)}$$ Lemma 7 : $f(1)^2=1$ Proof : Let $Q(x)$ be the assertion $f(xf(x))=x^2$ $Q(1) \implies f(f(1))=1$ Consider $Q(xf(x))$ We have, $f(xf(x)f(xf(x)))=(xf(x))^2$ That is, we have $f(x^3f(x))=x^2f(x)^2$ Setting $x=1$ here, we have $f(f(1))=f(1)^2$ Hence we have $$ \boxed {f(1)^2=1}$$ Lemma 8 : $f(yf(1))+f((y+2)f(-1))+2=0$ Proof : Consider $P(1,y)$ $$\implies f(f(y+1))=f(yf(1))+1$$Now consider $P(-1,y+2)$ $$\implies f(-f(y+1))=f((y+2)f(-1))+1$$ Now using $f$ is odd (Lemma 5), and adding the two equations, $$f(f(y+1))+f(-f(y+1))=f(yf(1))+f((y+2)f(-1))+2$$That is $$\boxed {f(yf(1))+f((y+2)f(-1))+2=0}$$ Main Proof : Since $f(1)^2=1$, we will form $2$ cases Also observe that by lemma 6, $f(2)=2f(1)$, Case 1 : $f(1)=1$ Since $f$ is odd, $f(-1)=-1$ Also, $f(2)=2$ Lemma 8 yields, $f(y)+f(-y-2)+2=0$ That is, $f(y+2)=f(y)+2$ Consider $P(2,y)$ and using $f(2x)=2f(x)$, $$f(2f(y+2))=f(yf(2))+4$$$$\implies 2f(f(y+2)) = f(2y)+4$$$$\implies f(f(y+2))=f(y)+2$$$$\implies f(f(y+2))=f(y+2)$$Usinf injectivity $f(y+2)=y+2$ That is $$\boxed{f(x)=x}$$This is a solution. Case 2 : $f(1)=-1$ This yields $f(2)=-2$ and $f(-1)=1$ Lemma 8 yields $f(-y)+f(y+2)+2=0$ That is $f(y+2)=f(y)-2$ Again consider $P(2,y)$ and use $f(2x)=2f(x)$ $$f(2f(y+2))=f(yf(2))+4$$$$\implies 2f(f(y+2))=-2f(y)+4$$$$\implies f(f(y+2))=-f(y+2)$$$$\implies f(f(y+2))=f(-(y+2))$$Using injectivity, $f(y+2)=-(y+2)$ That is $$\boxed {f(x)=-x}$$ So the only solutions are $\boxed{f(x) \equiv x}$ and $\boxed{f(x) \equiv -x}$

Similar to above. Denote by $P(x,y)$ the assertion. Note that $f(x)=x$ and $f(x)=-x$ are solutions.

Plugging in $f(x)=cx$ in the original equation, we get $f(x)=\pm x$ as the only solutions.

Orestis_Lignos wrote: Could someone check this? It is quite different from anyone else's solutions (I think) and I wonder whether there is something wrong with it. Let $P(x,y)$ be the given functional equation. We have that $P(0,x) \Rightarrow f(xf(0))=f(0)$, which if $f(0) \neq 0$ implies that $f$ is constant, which is easily seen to be a contradiction. Hence, $f(0)=0$. Therefore $$P(x,0) \Rightarrow f(xf(x))=x^2$$We now proceed with some Claims: Claim 1: $f(x)=0$ if and only if $x=0$ Proof: Let $a \in \mathbb{R}$ such that $f(a)=0$. Then, $a^2=f(af(a))=f(0)=0$, hence $a=0$ $\blacksquare$. Claim 2: $f(x)$ is injective. Proof: Suppose that there existed $A \neq B$ such that $f(A)=f(B)$. Then, $$P(B,A-B) \Rightarrow f(Bf(A))=f((A-B)f(B))+B^2,$$and since $f(Bf(A))=f(Bf(B))=B^2$, we obtain $f((A-B)f(B))=0$, which by Claim 1 implies $(A-B)f(B)=0$, hence $f(B)=f(A)=0$ which by Claim 1 implies $A=B=0$, a contradiction. Therefore, $f(x)$ is injective $\blacksquare$. Claim 3: $f$ is odd. Proof: Since $f$ is injective and $$f(xf(x))=x^2=(-x)^2=f(-xf(-x)),$$we have $xf(x)=-xf(-x)$ hence $f(-x)=-f(x)$ if $x \neq 0$, which along with $f(0)=0$ implies that $f$ is odd $\blacksquare$. Back to the problem, $x=1$ at $f(xf(x))=x^2$ implies $f(f(1))=1$, hence $f(1)^2=f(f(1)f(f(1)))=f(f(1))=1$, implying $f(1) \in \{1,-1 \}$. Case 1: $f(1)=1$. Since $f$ is odd $f(-1)=-1$. Now, $P(1,x) \Rightarrow f(f(x+1))=f(x)+1$. In addition, $$P(x,-1-x) \Rightarrow f(xf(-1))=f((-1-x)f(x))+x^2 \Rightarrow f((x+1)f(x))=f(x)+x^2.$$Note that $$P(x+1,-1) \Rightarrow f((x+1)f(x))=f(-f(x+1))+(x+1)^2=(x+1)^2-f(f(x+1))=(x+1)^2-(f(x)+1),$$therefore $$(x+1)^2-f(x)-1=f((x+1)f(x))=f(x)+x^2,$$which easily turns out to give $f(x)=x$, which evidently satisfies. Case 2: $f(1)=-1$. Since $f$ is odd $f(-1)=1$. Now, $P(1,x) \Rightarrow f(f(x+1))=1-f(x)$. In addition, $$P(x,-1-x) \Rightarrow f(xf(-1))=f((-1-x)f(x))+x^2 \Rightarrow f((x+1)f(x))=x^2-f(x).$$Note that $$P(x+1,-1) \Rightarrow f((x+1)f(x))=f(-f(x+1))+(x+1)^2=(x+1)^2-f(f(x+1))=(x+1)^2-(1-f(x)),$$therefore $$(x+1)^2+f(x)-1=f((x+1)f(x))=x^2-f(x),$$which easily turns out to give $f(x)=-x$, which evidently satisfies as well. Therefore, there are only two solutions, namely $f(x) \equiv x$ and $f(x) \equiv -x$. This is exactly my solution!

We claim that the only solutions are $f(x) = -x$ and $f(x) = x$, which clearly work. Now we prove that these are the only ones. $P(0,y): f(0) = f(yf(0)) \implies$ either $f$ is constant or $f(0)=0$. Since $f$ constant does not work, we have $f(0) = 0$. $P(x,0): f(xf(x)) = x^2$ and $P(x,-x): f(-xf(x)) = -x^2$ together implie that $f$ is surjective. Now we prove that $f$ is injective. First, take $k$ st $f(k)=0$. We have that: $P(k,0): f(kf(k)) = k^2 \implies k=0$ so $f$ is injective on zero. Now take $a,b$ st $f(a)=f(b)$. $P(x,a-x)$ and $P(x,b-x)$ implie $f(xf(a)) = f((a-x)f(x)) + x^2 = f(xf(b)) = f(b-x)f(x))$ so taking $x=a$ we have $f((b-a)f(a)) = 0$ so $b=a$. Due to injectivity and $P(x,-x), P(x,0)$; we conclude that $f(x) = -f(-x)$. Now take $c$ st $f(c) = 1$ (due to surjectivity). $P(c,0): f(cf(c)) = 1 = c^2$ and $P(1,0): f(f(1)) = 1$ so we get $f(1) = 1$ or $-1$ (so due to oddity $f(-1)$ is $-1$ or $1$). $P(1,x): f(f(x+1)) = f(xf(1)) + 1 \implies f(f(x)) = f((x-1)f(1)) + 1$ $P(-1,x+1): f(f(x)) = f((x+1))f(1)) -1$ so combining these two we get $f((x+1)f(1)) = f((x-1)f(1)) + 2 \implies f((x+2)f(1)) = f(xf(1)) + 2$ So now we apply $P(x+2,-x-1): f((x+2)f(1)) = -f((x+1)f(x+2)) + (x+2)^2 = f(f(x+1)) - (x+1)^2 +(x+2)^2 = -f(xf(1)) + 2x + 2$ so we conclude $f(xf(1)) = x$. Since we have $f(1) = 1$ or $-1$, we're done.

Let $P(x,y)$ denote the given assertion. And let $P(x,y)\circ P(u,v)$ denote the combination of $P(x,y)$ and $P(u,v).$ Trivially we have $f(x)=0$ iff $x=0.$We now prove a lemma. Lemma: $f$ is injective. Proof. If $f(m)=f(n)$ for some $m,n,$ then $m^2=f(mf(m))=f(mf(n))=m^2+f((n-m)f(m))\implies f((n-m)f(m))=0 \implies m=n.$ $\blacksquare$ $P(x,0)\circ P(-x,0)\implies f(xf(x))=f(-xf(-x))\implies f(-x)=-f(x).$ $P(x,x) \circ P(x,-2x)\implies f(2xf(x))=f(xf(2x))\implies f(2x)=2f(x).$ $P(1,x)\implies f(f(x+1))=f(x)+1.$ $P(1,0)\circ P(1,1)\implies f(1)=1\text{ or } f(1)=-1.$ $P(x,-x-1)\circ P(x+1,-1)\implies$ $f(x)= \begin{cases} x &\text{if }f(1)=1, \\ -x &\text{if }f(1)=-1,\end{cases}$ both of which fit.

:skull: Let $P(x,y)$ denote the assertion. $P(0,y)$ gives $f(0)=f(yf(0)).$ If $f(0)$ is not zero, then $yf(0)$ is surjective, which means $f(x)$ is constant. However, back in the original equation, this implies $x^2=0$ so $f(x)$ is not constant, which means $f(0)=0.$ Now, $P(x,0)$ gives $f(xf(x))=x^2$ and $P(x,-x)$ gives $f(-xf(x))=-x^2.$ $~$ Suppose $f(x)=f(y)$ then $f(xf(x))=f(xf(y))=x^2$ for all $y.$ However, $f(xf(y))=f((y-x)f(x))+x^2$ which implies $f((y-x)f(x))=0.$ Note that if $f(x)=0$ then $0=f(xf(x))=x^2$ which implies $x=0$, but this means $f$ is injective. $~$ Now, $P(1,y)$ implies $f(f(1+y))=f(yf(1))+1$, but $f(f(1))=1$ so $f(1)f(f(1))=f(1)$ which implies $f(f(1)f(f(1)))=f(f(1))=f(1)^2$ so $f(1)^2=1.$ Now, WLOG assume $f(1)=1.$ $P(x,x)$ gives $f(xf(2x))=2x^2=f(\sqrt{2}xf(\sqrt{2}x))$ so $f(2x)=\sqrt{2}f(\sqrt{2}x)$ so $f(2x)=2f(x).$ In particular, $f(2)=2$ and $P(2,x)$ implies $2f(f(2+y))=2f(y)+4$ which implies $f(y+2)=f(y)+2$ $~$ Since $f(f(1+y))=f(y)+1$ This implies $f(f(2+y))=f(y)+2=f(y+1)+1$ which implies $f(1+y)=f(y)+1$ so $f(1+y)=1+y$ so $f(y)=y.$ The other case gives $f(y)=-y$ and both work.

IMO SL 2009 A7 wrote: Find all functions $f$ from the set of real numbers into the set of real numbers which satisfy for all $x$, $y$ the identity \[ f\left(xf(x+y)\right) = f\left(yf(x)\right) +x^2\] Let $P(x,y)$ be assertion of the given equation. $P(0,y)\implies f(yf(0))=f(0)$. If $f(0)\ne 0$, then $f(x)\equiv c$ for all $x$ which doesn't work. So, $f(0)=0$. $P(x,0) \implies f(xf(x))=x^2$. Assume there is $t\ne 0$ such that $f(t)=0$. $P(t,0)\implies t^2=0$ which is contradiction. So, $f$ is injective at $0$. Assume there are $a\ne b$ such that $f(a)=f(b)$. Obviously, $ab\ne 0$. $P(a,b-a)\implies a^2=f(af(a))=f(af(b))=f((b-a)f(a))+a^2 \implies f((b-a)f(a))=0 \implies (b-a)f(a)=0$. Since $a\ne 0,$ we get $a=b$, which is contra. So, there are no such $a,b$, which means $f$ is injective. $f(xf(x))=x^2=(-x)^2=f(-xf(-x))\implies xf(x)=-xf(-x)\implies f(-x)=-f(x)$. $P(x,x)\implies f(xf(2x))=2x^2=(\sqrt2 x)^2=f(\sqrt2 xf(\sqrt2 x)) \implies xf(2x)= \sqrt2 xf(\sqrt2 x) \implies f(\sqrt2 z)=\sqrt2 f(z)\implies f(2x)=\sqrt2 f(\sqrt2 x)=2f(x)\implies f(2x)=2f(x)$. $P(2,2x-2)\implies f(2f(2x))=f((2x-2)f(2))+4\implies f(f(x))=f((x-1)f(1))+1$. Let $Q(x)$ be assertion of $f(f(x))=f((x-1)f(1))+1$. $Q(2x)\implies f((2x-1)f(1))+1=f(f(2x))=2f(f(x))=2f((x-1)f(1))+2=f((2x-2)f(1))+2$. Rewrite $(2x-1)f(1)=r \implies f(r+f(1))=f(r)+1$ for all $r\in\mathbb{R}$. Now take $r=(s-1)f(1) \implies f(sf(1))=f((s-1)f(1))+1=f(f(s))\implies sf(1)=f(s) \implies f(x)=cx$ for all $x$. Since $f(f(1))=1$, we get $c=1$ or $c=-1$. So we get $2$ solutions $f(x)=x$ and $f(x)=-x$ and both indeed work.

Let $x=0,$ then $f(0)=f(yf(0)).$ If $f(0)\neq 0,$ then $f(y)\equiv C$ where ${C}$ is a constant$.$ Then $0=0+x^2,x\in\mathbb R,$ which is obviously wrong$.$ Therefore $f(0)=0.$ Let $y=0,$ then $f(xf(x))=x^2\cdots (*).$ We have $f(x)=0\Leftrightarrow x=0.$ If $\exists a,b\in\mathbb R,a\neq b,f(a)=f(b),$ then $$a^2\stackrel{(*)}{=}f(af(a))=f(af(b))=f((b-a)f(a))+a^2\Rightarrow f((b-a)f(a))=0$$$$\Rightarrow (b-a)f(a)=0\Rightarrow f(a)=0\Rightarrow a=b=0$$which is not true$.$ Therefore for $\forall a,b\in\mathbb R,f(a)\neq f(b).$ As for $x\neq 0,$ $$f(xf(x))\stackrel{(*)}{=}x^2=(-x)^2\stackrel{(*)}{=}f(-xf(-x))\Rightarrow xf(x)=-xf(-x)\Rightarrow f(x)=-f(-x).$$Let $x=y\neq 0$, we have $f(xf(2x))=f(xf(x))+x^2\stackrel{(*)}{=}2x^2=(\sqrt 2x)^2\stackrel{(*)}{=}f(\sqrt 2xf(\sqrt 2x)).$ Then $$xf(2x)=\sqrt 2xf(\sqrt 2x)\Rightarrow f(2x)=\sqrt 2f(\sqrt 2x)\Rightarrow f(2x)=\sqrt 2f(\sqrt 2x)=2f(x).$$As $f(f(1))\stackrel{(*)}{=}1,f^2(1)\stackrel{(*)}{=}f(f(1)f(f(1)))=f(f(1))=1\Rightarrow f(1)=\pm 1.$ If $f(1)=1.$ Then $f(2)=2f(1)=2.$ Let $x=1,f(f(y+1))=f(yf(1))+1=f(y)+1.$ Let $x=2,$ $$2f(f(y+2))=f(2f(y+2))=f(yf(2))+4=f(2y)+4=2f(y)+4\Rightarrow f(f(y+2))=f(y)+2.$$$$\therefore f(y)+2=f(f(y+2))=f(y+1)+1\Rightarrow f(y+1)=f(y)+1=f(f(y+1))$$$$\Rightarrow f(y+1)=y+1\Rightarrow f(x)=x,\forall x\in\mathbb R.$$If $f(1)=-1.$ Then $f(2)=2f(1)=-2.$ Let $x=1,f(f(y+1))=f(yf(1))+1=f(-y)+1.$ Let $x=2,$ $$2f(f(y+2))=f(2f(y+2))=f(yf(2))+4=f(-2y)+4=2f(-y)+4\Rightarrow f(f(y+2))=f(-y)+2.$$$$\therefore f(-y)+2=f(f(y+2))=f(-(y+1))+1\Rightarrow f(-y-1)=f(-y)+1=f(f(y+1))$$$$\Rightarrow f(y+1)=-y-1\Rightarrow f(x)=-x,\forall x\in\mathbb R.$$Therefore $\boxed{f(x)=\pm x}.\blacksquare$

Let $P(x,y)$ denote the assertion in the problem. The only solutions are $f(x) \equiv x$ and $f(x) \equiv -x$, which work. Claim 1: $f(x)=0$ if and only if $x=0$. Proof: If $f(0) \ne 0$, then $P(0,x)$ gives $f(0)=f(xf(0))$, so $f$ is constant, a contradiction. If $f(x)=0$, then $P(x,0)$ gives $x^2=0$, so $x=0$. $\square$ Claim 2: $f$ is bijective. Proof: $P(x,0)$ gives $f(xf(x))=x^2$ and $P(x,-x)$ gives $f(-xf(x))=-x^2$, so $f$ is surjective. To prove that $f$ is injective, suppose that $f(a)=f(b)$ for real numbers $a$ and $b$. Then, $P(a,b-a)$ implies $f(af(b))=f((b-a)f(a))+a^2$. We also have $f(af(b))=f(af(a))=a^2$, so $f((b-a)f(a))=0$, which implies $a=0$ or $a=b$ by Claim 1. If $a=0$, then Claim 1 implies $a=b$, as desired. $\square$ Claim 3: $f$ is odd. Proof: Plug in $P(x,0)$ and $P(-x,0)$ to get $f(xf(x))=f(-xf(-x))=x^2$, which gives $f(-x)=-f(x)$. $\square$ Claim 4: $f$ is additive. Proof: By $P(x,\tfrac{y}{f(x)})$ and $P(x,-\tfrac{y}{f(x)}-2x))$ for nonzero $x$, we have \[f(xf(x+\tfrac{y}{f(x)}))=f(y)+x^2=f((\tfrac{y}{f(x)}+2x)f(x))-x^2 \implies f(y+2xf(x))=f(y)+2x^2.\]Setting $y=0$ gives $f(2xf(x))=2x^2$, so if $z=2xf(x)$, then $f(y+z)=f(y)+f(z)$. Replacing $y$ with $-y$ and using oddness gives $f(y-z)=f(y)+f(-z)$. However, notice that \[\mathbb{R}=f^{-1}(\mathbb{R})=f^{-1}(\{2x^2 \colon x \in \mathbb{R}\}) \cup f^{-1}(\{-2x^2 \colon x \in \mathbb{R}\})=\{2xf(x) \colon x \in \mathbb{R}\} \cup \{-2xf(x) \colon x \in \mathbb{R}\},\]so the possible values of $z$ and $-z$ comprise all of $\mathbb{R}$, which means $f$ is additive. $\square$ Thus, $P(x,y)$ gives \[f(yf(x))+x^2=f(xf(x)+xf(y))=f(xf(x))+f(xf(y)) \implies f(xf(y))=f(yf(x)) \implies xf(y)=yf(x),\]so $f(x) \equiv cx$ for some real number $c$. We also have $f(xf(x))=x^2$, so $c \in \{1,-1\}$, as desired. $\square$

Let $P(x,y)$ denote the assertion in the problem. The only solution that satisfies are $f(x)=x \ \forall \ x \in \mathbb{R}$ and $f(x)=-x \ \forall \ x \in \mathbb{R}$. Claim 1. $f(x)=0$ if and only if $x=0$. Proof. Note that $f$ is not constant, from $P(0,x)$ we have $f(0)=f(xf(0))$. If $f(0)\neq 0$, then $xf(0)$ can be any value in $\mathbb{R}$. Therefore $f$ is constant, contradiction. Hence $f(0)=0$. If $f(x)=0$ for some $x \in \mathbb{R}$, from $P(x,-x)$ we can derive $x^2=0$, implying $x=0$. Hence $f(x)=0$ if and only if $x=0$. Claim 2. $f(xf(x))=x^2 \ \forall \ x \in \mathbb{R}$. Proof. The proof comes immediately from $P(x,0)$. Claim 3. $f$ is injective. Proof. If there exist $a,b \in \mathbb{R}$ such that $f(a)=f(b)$, then from $P(b,a-b)$ we have $$f(bf(a))=f((a-b)f(b))+b^2$$From claim $2$, we have $f(bf(a))=f(bf(b))=b^2$. Therefore $f((a-b)f(b))=0$. From claim $1$, we can conclude $(a-b)f(b)=0$. If $f(b)=0$ then we have $a=b=0$ from claim $1$. If $f(b)\neq 0$, then $a-b=0$ giving $a=b$. Hence $f$ is injective. Claim 4. $f(2x)=2f(x) \ \forall \ x \in \mathbb{R}$. Proof. From $P(x,x)$ we have $$f(xf(2x))=f(xf(x))+x^2=2x^2$$From claim $2$, we have $f(\sqrt{2}xf(\sqrt{2}x))=(\sqrt{2}x)^2=2x^2$. Since $f$ is injective, we can conclude $$xf(2x)=\sqrt{2}xf(\sqrt{2}x)$$Obviously the claim is true for $x=0$, if $x\neq 0$ then we have $f(2x)=\sqrt{2}f(\sqrt{2}x)$. Substituting $x=\frac{\sqrt{2}}{2}x$ gives $f(\sqrt{2}x) =\sqrt{2}f(x)$. Therefore, $$f(2x)=\sqrt{2}f(\sqrt{2}x)=2f(x)$$As desired. From claim $2$, we have $f(f(1))=1$. Substituting $x=f(1)$ to claim $2$ gives $f(1)^2=1$. Case 1. $f(1)=1$ From claim $2$ we have $f(-f(-1))=1$, since $f$ is injective we have $f(-1)=-1$. From $P(1,x+1)$ we have $$f(f(x+2))=f(x+1)+1$$From $P(2,x)$ and claim $4$, $$f(2f(x+2))=f(2x)+4 \Rightarrow 2f(f(x+2))=2f(x)+4 \Rightarrow f(f(x+2))=f(x)+2$$From those two equation we have $$f(x+1)=f(x)+1 \Rightarrow f(x+2)=f(x+1)+1$$Substituting this back to the first equation we have $f(f(x+2))=f(x+2)$, since $f$ is injective we have $f(x)=x \ \forall \ x \in \mathbb{R}$. Case 2. $f(1)=-1$ Since this case is very similar to the former, I will leave it to the reader. This case gives us $f(x)=-x \ \forall \ x \in \mathbb{R}$. Having all cases exhausted, we are done.

2009 A7 wrote: Find all functions $f$ from the set of real numbers into the set of real numbers which satisfy for all $x$, $y$ the identity\[ f\left(xf(x+y)\right) = f\left(yf(x)\right) +x^2\] Fun problem! We claim that $f \equiv id,-id$ Claim 1: $f(0)=0$

Claim 2: $f$ is injective at $0$

Claim 3: $f$ is bijective

Claim 4: $f$ is odd

Claim 5: $f$ is solution, so is $-f$

Claim 6: $f(1)=1$

Claim 7: $f(2x)=2f(x)$

Claim 8 $f(y+2)=f(y)+2$

Claim 9: $f$ is idempotent.

Hence the only solutions are indeed $f \equiv x,-x$

First A7! If $f(0)\neq 0$ then $(x,y)=\left(0,\frac x{f(0)}\right)\Rightarrow f(x)=f(0)$, so $f$ is a constant function but this is clearly impossible. It follows that $f(0)=0$. Now if we put $y=0$ we find that $f(xf(x))=x^2$. Suppose $f(t)=0$ for some non zero real $t$, then $t^2=f(tf(t))=0$ a contradiction. Thus $f(x)=0$ iff $x=0$. Now we wish to show that $f$ is injective. So let us assume $f(a)=f(b)\neq 0$, then $$(x,y)=(b,a-b)\Rightarrow f((a-b)f(b))+b^2=f(bf(a))=f(bf(b))=b^2.$$So $f((a-b)f(b))=0$, and it follows from our earlier observation that $(a-b)f(b)=0$, thus $a=b$ as desired. $(x,y)=(1,0)\Rightarrow f(f(1))=1$ and so $(x,y)=(f(1),0)\Rightarrow 1=f(1)^2$. We will suppose WLOG that $f(1)=1$. Now $(x,y)=(1,-1)$ also gives $f(-1)=-1$. If we set $(x,y)$ to be $(1,y-1)$ and $(-1,-y+1)$ then we find $$\boxed{f(f(y))=f(y-1)+1}=f(-f(-y)).$$So by injectivity, $f(y)=-f(-y)$ and so $f$ is odd. For the final blow, \begin{align*} & (x,y)=(1-x,x)\Rightarrow f(1-x)=f(xf(1-x))+(1-x)^2\Rightarrow f(x-1)=f(xf(x-1))-(x-1)^2 \\ & (x,y)=(x,-1)\Rightarrow f(xf(x-1))=f(-f(x))+x^2=-f(x-1)-1+x^2.\end{align*}Thus $f(x-1)=-f(x-1)-1+x^2-(x-1)^2$, it follows that $f(x-1)=x-1$. So $f(x)=x$ for all real $x$. It is easy to check that this is a solution. If we had chosen $f(1)=-1$ then we would find $f(x)=-x$ which is also a solution.

There is an alternative way to end the problem after after we proved the injectivity, $-f(x)=f(-x)$ and $f(2x)=2f(x)$ Put $P(x, x+y): $ $$f(xf(2x+y))=f((x+y)f(x))+x^2;$$$P(2x,y):$ $$f(2xf(2x+y))=f(yf(2x))+4x^2$$note that $$f(yf(2x))+4x^2=2f(yf(x))+4x^2=2f((x+y)f(x))+2x^2$$$$f(yf(x))+x^2=f((x+y)f(x))$$but LHS is equal to $f(xf(x+y))$ So due to injectivity $xf(x+y)=(x+y)f(x)$, then $f(x)/x=cons$, implying $f(x)=x$ or $f(x)=-x.$

The answer is $x$ and $-x$. Let $P(x,y)$ be the assertion. $P(0,y)$ implies $f$ is constant or $f(0)=0$, $f$ is not constant so $f(0)=0$. $P(x,0)$ means $f(xf(x))=x^2$. $f$ is injective at $0$ because if $c$ is such that $c=0$, then $f(0)=c^2$, so $c=0$. If $f(x+y)=f(x)$, $f(yf(x))+x^2=f(xf(x+y))=f(xf(x))=x^2$, so $f(yf(x))=0$ and either $y=0$ or $f(x)=0$, but when $f(x)=0$ then $x=0$ and $f(y)\neq 0$ unless $y=0$. Thus $f$ is injective. In particular, since $f(xf(x))=x^2=f(-xf(-x))$, then $xf(x)=-xf(-x)$ and $f(x)=-f(-x)$. $P(-x-y,x)$ gives, after taking negatives out, $f((x+y)f(y))=-f(xf(x+y))+(x+y)^2=-f(yf(x))+y(y+2x)$. Plugging in $y=-2x$ and taking out negatives gives $f(xf(2x))=f(2xf(x))$, so $xf(2x)=2xf(x)$ and $f(2x)=2f(x)$. $P(x+y,x)$ gives $f((x+y)f(2x+y))=f(xf(x+y))+(x+y)^2=f(yf(x))+(x+y)^2+x^2$. Thus if we let $a=y$ and $b=2x+y$, $f\left(\frac{a+b}{2}f(b)\right)-f\left(\frac{a+b}{2}f(a)\right)=\frac{b^2-a^2}{2}$ for all $a,b\in \mathbb{R}$. Thus $f((a+b)f(b))-f((a+b)f(a))=b^2-a^2$. This means $f((x+y)f(y))-f((x+y)f(x))=y^2-x^2$, but because of the equation $f((x+y)f(y))=(x+y)^2-x^2-f(yf(x))$ and its symmetric relation, $f(yf(x))=f(xf(y))$, so $yf(x)=xf(y)$. Thus $\frac{f(x)}{x}$ is constant for all $x$, and we can get $x$ or $-x$ as the only solutions.

My first A7 Denote by $P(x, y)$ the given assertion. Claim 1: $f(0)=0$ Proof: Suppose $f(0) \neq 0$ $P(0, \frac{1}{f(0)}): f(0)=f(1)$ $P(0, 1): f(0)=f(f(0))$ $P(1,0): f(f(1))=f(0)+1$ Plugging the first two relations into the third one, we get a contradiction. So $f(0)=0$ and moreover, if $f(x)=0$ for some $x \in \mathbb{R}$ then $x=0$ $\square$ Claim 2: $f$ is injective Proof: Suppose we can find $a, b \in \mathbb{R}$ such that $a < b$ and $f(a)=f(b)$ $P(a, b-a): f(af(b))=f((b-a)f(a))+a^2$ But $f(af(b))=f(af(a))=a^2$ from $P(a,0)$ therefore $f((b-a)f(a))=0$ and using claim 1, this leads to $a=b$, contradiction. So $f$ is injective. $\square$ Notice that $P(\sqrt{a}, 0)$ and $P(\sqrt{a},-\sqrt{a})$ give that $f$ is surjective. Therefore we may find $c \in \mathbb{R}$ such that $f(c)=1$ $P(c,0): f(cf(c))=c^2$ $\Rightarrow c^2=1$ $\Rightarrow c \pm 1$ But it’s easy to see that if $f$ is a solution, so is $-f$ therefore we can assume WLOG that $c=1$. So $f(1)=1$ Similarly, denote by $e$ the number for which $f(e)=-1$. By $P(e, -e): f(-ef(e))=-e^2$ $\Rightarrow e^2=1$ but from injectivity $e=-1$ therefore $f(-1)=-1$. Claim 3: $f$ is odd. Proof: $P(1, x):f(f(x+1))=f(x)+1$ $P(-1, x+1):f(-f(x))=f(-x-1)+1$ $P(1, -x-1): f(f(-x))=f(-x-1)+1$ $\Rightarrow f(-x)=-f(x)$ $\square$ Since $f$ is odd, $P(-1,x+1): -f(f(x))=-f(x+1)+1$ $\Rightarrow f(f(x))+1=f(x+1)=f(f(x)+1)$ so, by injectivity, $f(x)=x$. Therefore, the only solutions are $f(x)=x$ and $f(x)=-x$ $\blacksquare$