Determine all $P(x) \in \mathbb R[x]$ such that $P(x^2) + x\big(mP(x) + nP(-x)\big) = \big(P(x)\big)^2 + (m - n)x^2$, $\forall x \in \mathbb R$. 
Determine all polynomials $P(x) \in \mathbb R[x]$ knowing that $$ P(x^2) + x\big(mP(x) + nP(-x)\big) = \big(P(x)\big)^2 + (m - n)x^2, \forall x \in \mathbb R\,.$$

Replacing $x$ by $-x$, we have that $$P(x^2) - x[mP(-x) + nP(x)] = [P(-x)]^2 + (m - n)x^2, \forall x \in \mathbb R$$
Subtracting the second equation from the first, $$x(m + n)[P(x) + P(-x)] = P(x)^2 - P(-x)^2, \forall x \in \mathbb R$$
$$ \iff \left[ \begin{align} P(x) + P(-x) &= 0\\ P(x) - P(-x) &= x(m + n) \end{align} \right.$$
In the case of $P(x) +P(-x)=0$, we have that $$P(x^2) + (m - n)xP(x) = [P(x)]^2 + (m - n)x^2$$
In the case of $P(x) - P(-x) = x(m + n)$, we have that $$P(x^2) + x(m + n)[P(x) -nx] = [P(x)]^2 + (m - n)x^2$$
That neither of which can I answer is not unexpected.

We conjecture that for odd degree $s\geq 3$ and above, the only polynomials that work are
$$P(x)=x^s+\frac{m-n}{2}x\text{ for }(m - n) (-2 + m - n)=0$$
while for even degree $s>3$ the only polynomials that work are
$$P(x)=x^s+\frac{m+n}{2}x\text{ for }(m - 3 n) (-2 + m + n)=0$$
In order to prove this, let
$$P(x)=\sum_{i=0}^s a_i x^i$$
where $a_s\neq 0$ and $s\geq 3$. Then by the equation given
$$0=P(x^2) + x (m P(x) + n P(-x))-P(x)^2-(m-n)x^2$$
$$=\sum_{i=0}^s a_i x^{2i}+\sum_{i=0}^s ma_i x^{i+1}+\sum_{i=0}^s na_i (-1)^ix^{i+1}-\sum_{i=0}^s\sum_{j=0}^sa_ia_jx^{i+j}-(m-n)x^2$$
We compute the highest coefficients first. Obviously, the highest power of $x$ is at most $2s$. This coefficient is
$$0=a_s-a_s^2\Rightarrow 0=a_s-1$$
This implies $a_s=1$. For $x^{2s-1}$, we have
$$0=-2a_{s-1}a_s=-2a_{s-1}$$
This implies $a_{s-1}=0$. For $x^{2s-2}$ we have
$$0=a_{s-1}-2a_{s-2}a_s-a_{s-1}^2=-2a_{s-2}$$
Again, this implies $a_{s-2}=0$. By induction, this pattern holds until $a_1$. It is this induction step that we use the fact that $s\geq 3$, else the pattern would break from $(m-n)x^2$. That is, we know
$$a_{s-1}=a_{s-2}=\cdots a_3=a_2=0$$
Then the polynomial is of the form
$$P(x)=x^s+ ax+b$$
Putting this into the equation gives us
$$0=(b-b^2)+x b (-2 a + m + n)+x^2(1-a) (a - m + n)+x^s(-2b)+x^{s+1}(m+n(-1)^s-2a)$$
Obviously, $b=0$. Then the equation becomes 
$$0=x^2(1-a) (a - m + n)+x^{s+1}(m+n(-1)^s-2a)$$
Then 
$$a=\frac{m+n(-1)^s}{2}$$
This gives us the first part of our conjecture. For the second part
$$0=(1-a) (a - m + n)=(2-m+n(-1)^s)(m+n(-1)^s-2m+2n)$$
Thus, either
$$2-m+n(-1)^s=0$$
or
$$m+n(-1)^s-2m+2n=0$$
For degree $0$, we have
$$P(x)=a$$
for some $a\in\mathbb{R}$. This implies
$$0=(a-a^2)+ax(m+n)+x^2(n-m)$$
Since this holds for all $x$, we know $a\in\{0,1\}$ and $m-n=0$. Without knowing more about $m$ and $n$, this is the best we can do. Now, let
$$P(x)=ax+b$$
for $a\neq 0$. Then
$$0=b(1-b)+xb (-2 a + m + n)+x^2(1 - a) (a - m + n)$$
This implies
$$b\in\{0,1\}$$
but other than than we can't say anything else without more restrictions on $m$ and $n$. Note that $(a,b)=(1,0)$ will work no matter what $n$ and $m$ are. For quadratics, let
$$P(x)=ax^2+bx+c$$
for $a\neq 0$. Then
$$0=c - c^2 + (-2 b c + c m + c n) x + (b - b^2 - 2 a c - m + b m + n - 
    b n) x^2 + (-2 a b + a m + a n) x^3 + (a - a^2) x^4$$
Since $a\neq 0$, this implies $a=1$. Then
$$0=c - c^2 + (-2 b c + c m + c n) x + (b - b^2 - 2 c - m + b m + n - 
    b n) x^2 + (-2 b + m + n) x^3$$
which implies
$$b=\frac{m+n}{2}\text{ and }c\in\{0,1\}$$
Then
$$P(x)=x^2+\frac{m+n}{2}x+c\text{ for }-2 c + \frac{1}{4} (m - 3 n) (-2 + m + n)=0\text{ with }c\in\{0,1\}$$
Having checked all degrees, let us discuss the results. First, if $m,n$ are arbitrary, then the only polynomial that works is $P(x)=x$. Second, we can list out all equations of $m$ and $n$ that give infinite solutions. We have
$$m - n=0$$
$$m - 3n=0$$
$$m+n=2$$
$$m-n=2$$
That is, if $m$ and $n$ do not satisfy one of these equations, then $P(x)$ has only finite solutions. The final equation which also gives a finite solution set is
$$-8 - 2 m + m^2 + 6 n - 2 m n - 3 n^2=0$$
which creates the quadratic
$$P(x)=x^2 + \frac{m + n}{2} x + 1$$
