Can we obtain $f(y+x)=y+f(x)$ from $f(x^2+f(x)^2+x)=f(x)^2+x^2+f(x)$? 
$\mathbb Z^+$ is the set of positive integers. Find all functions $f:\mathbb{Z}^+\rightarrow \mathbb{Z}^+$ such that
  $$f(m^2+f(n))=f(m)^2+n\quad(\clubsuit)$$

Let $P(x,y)$ be the assertion: $f(x^2+f(y))=f(x)^2+y \; \forall x,y \in \mathbb{Z}^+.$
$P(x,x)$ gives us $f(x^2+f(x))=f(x)^2+x$.
$P(x,x^2+f(x))$ gives us $f(x^2+f(x)^2+x)=f(x)^2+x^2+f(x)$. 
Can we obtain $f(y+x)=y+f(x)$ from $f(x^2+f(x)^2+x)=f(x)^2+x^2+f(x)$ ?

Given
$$
f: \mathbb{Z} \rightarrow \mathbb{Z} \wedge f(m^2 + f(n)) = f(m)^2 + n.
$$
For a fixed $m$ and $n=x$ we get
$$
f(m^2 + f(x)) = f(m)^2 + x.
$$
In general we can write
$$
f(x) = \sum_k a_k x^k.
$$
Whence
$$
f'(m^2 + f(x)) f'(x) = 1.
$$
Thus $f'(x) = 1$.
So we obtain
$$
f(x) = a + x.
$$
Putting it back we get
$$
a + [ m^2 + a + n ] = a^2 + 2 a m + m^2 + n \Rightarrow 2a = a^2 + 2 a m.
$$
Whence
$$
a = 0.
$$
So the general solutions is
$$
f(x) = x.
$$
