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