# Is there any neat way to show $T$ is $\mathbb R$-linear?

Let $T: \mathbb R \to \mathbb R$ be the map which satisfies the following functional equation $T(x^2+T(y))=y+T(x)^2$ $\forall x,y \in \mathbb R$ Is there any neat way to show that $T$ is $\mathbb R$-linear?

After substituting some "special values" and doing a little bit algebraic manipulation it turned out that identity map on $\mathbb R$ is the only map which satisfies the above functional equation so $T$ must be $\mathbb R$ -linear. But Is there any other way (without computing all possible choices for $T$) to show that $T$ is $\mathbb R$-linear?

As we can vary $y$ arbitrarily, we see that $T$ is surjective. Note that $T$ must also be injective (hence bijective), for if $T(y_1)=T(y_2)$ then $$y_1 = T(42^2+T(y_1))-T(42)^2=T(42^2+T(y_2))-T(42)^2=y_2.$$ Next, $$T(-x)^2=T((-x)^2+T(42))-42=T(x^2+T(42))-42=T(x)^2$$ hence $T(-x)=\pm T(x)$ and by injectivity $$\tag1T(-x)=-T(x)\qquad \text{for all x\ne 0}.$$ By $(1)$ and injectivity we see that $T(x)\ne 0$ for all $x\ne 0$. Hence $$\tag2T(0)=0.$$ (Which allows us to extend $(1)$ to all $x$). Now we find $T(x^2+T(0))=0+T(x)^2$, i.e., $$\tag3 T(x^2)=T(x)^2$$ and hence $$\tag4T(x)\lessgtr0\iff x\lessgtr0.$$ Then for any $y<0$, we can let $x=\sqrt{-T(y)}$ and find $$0=T(0)=T(x^2+T(y))=y+T(x)^2=y+T(x^2)=y+T(-T(y))=y-T(T(y))$$ so that $T(T(y))=y$ if $y<0$. Using $(1)$, this also holds for $y>0$, in other words, $T$ is an involution: $$\tag5T(T(x))=x\qquad\text{for all x\in\mathbb R}.$$ Given $a,b\in \mathbb R$ with $a\ge 0$, pick $x=\sqrt a$, $y=T(b)$. Then $$\tag6T(a+b)=T(x^2+T(y))=y+T(x)^2=y+T(x^2)=T(b)+T(a)$$ (using $(1)$ this also holds if $a<0$) so that $T$ is additive. Using induction, $(6)$ implies that $$T(nx)=nT(x)\qquad\text{for all x\in\mathbb R, n\in\mathbb N_0}$$ then by $(1)$ $$T(nx)=nT(x)\qquad\text{for all x\in\mathbb R, n\in\mathbb Z}$$ and finally $$\tag7 T(qx)=qT(x)\qquad\text{for all x\in\mathbb R, q\in\mathbb Q}.$$ Combining $(4)$ and $(6)$ we find that $T$ is strictly increasing: If $y>x$ then $$\tag8 T(y)=T(x+(y-x))=T(x)+T(y-x)>T(x).$$ Together with $(7)$ this implies $$T(x)=xT(1)\qquad\text{for all x\in\mathbb R}$$ which is the desired result: $T$ is $\mathbb R$-linear.
In fact, the only increasing linear involution is the identity so that my attempt to show $\mathbb R$-linearity without finding all $T$ failed ...