Is a function $f$ satisfying $f(x+1) = f(x) + 1$ and $f(x^2) = (f(x))^2$ odd or even? The problem statement, all variables and given/known data

1) $f(x+1)=f(x)+1$
  2) $f(x^2) =(f(x))^2$

Let a function $f \colon \mathbb{R} \to \mathbb{R}$ satisfy the above statements. Then prove whether the fuction is odd or even.
The attempt at a solution:
using 2) we get
a) $f(0) = 0,1$ and
b) $f(1) = 0,1$
putting $x = 0$ in the 1st equation we get $f(0) = 0$ and $f(1) = 1$. From this we can prove $f(-1) = -1$ and for integers we get $f(-x) = -f(x)$. But how to prove for all real $x$?
 A: Proposition. $f(x)=x$ for all $x\in\mathbb R$.
Proof. Let $g(x)=x-f(x)$, $Y=\{g(x)\mid x\in\mathbb R\}$ and $s=\sup Y\in\mathbb R\cup\{+\infty\}$.
If $0\le x <1$, we find from  $(2)$ that $ f(x)=f(\sqrt x^2)=f(\sqrt x)^2\ge 0$ and hence $g(x)<1$.
By $(1)$, $g$ is periodic with period $1$ so that we have $g(x)<1$ for all $x\in\mathbb R$ and hence $s\le 1$ (especially, $s$ is finite).
As an intermezzo, we show a little

Lemma. If $y\in Y$ and $a\in\mathbb R$, then there exists $x\in[a,a+1)$ such that $(2x-y)y\in Y$.
Proof:
Since $g$ is periodic with period $1$,  there exists some $x\in[a,a+1)$ with $g(x)=y$. We
compute $(2x-y)y=(x+f(x))(x-f(x))=x^2-f(x)^2=x^2-f(x^2)=g(x^2)\in Y$. $_\square$

Back to the proof of the proposition.
Assume $Y\ne \{0\}$.
Let $y\in Y\setminus\{0\}$.
If $y>0$ then immediately $s\ge y>0$.
If $y<0$, let $a=\frac y2-1$ in the lemma and obtain $2x-y<0$, hence again $s\ge (2x-y)y>0$.
Therefore, we have $0<s\le 1$.
Select $y\in Y$ with $y>\frac s2>0$ and let  $a=1+\frac y2$ in the lemma, we obtain $2x-y\ge 2$, i.e. the contradiction $s\ge (2x-y)y\ge2y>s$.
We conclude that $Y=\{0\}$, i.e. the proposition. $_\square$
A: Hint:from  second condition 
$$\begin{cases}
f((-x)^2)=f(x^2)=(f(-x))^2 \\
f(x^2)=(f(x))^2 \\
\end{cases}\color{red}{\Rightarrow } 0=f(x^2)-f(x^2)=(f(x))^2-(f(-x))^2\color{red}{\Rightarrow } $$ $$(f(-x))^2=(f(x))^2\to f(x)=\pm f(-x)$$$$\color{red}{\Rightarrow }\begin{cases}
f(x)=f(-x)  \color{green}{(1)}\\
f(x)=-f(-x)  \color{green }{(2)}\\
\end{cases}$$  if we let  $\color{green}{(1)}$ be true but there are counter example like $f(x)=x^2$  such that  don't satisfy in $f(x+1)=f(x)+1$ 
and if  we let  $\color{green}{(2)}$ be true but there is odd function like $f(x)=x^3$ such that  don't satisfy in $f(x+1)=f(x)+1$ 
