Computing $m-n$ 
When $P(x+4)$ is divided by $P(x)$, the remainder is $3x+m$. When
  $P(x)$ is divided by $P(x+4)$, the remainder is $nx-6$. Compute $m-n$

Here I wrote down the equations as follows
$$P(x+4) = P(x)Q_1(x)+3x+m$$
$$P(x) = P(x+4)Q_2(x) + nx-6$$
However, these will not give me anything useful. Could you assist me with what I'm missing?
Regards

firstly, we write :
$m=P(x+4)-P(x)Q_1(x)-3x,$
$n=[P(x)-P(x+4)Q_2(x)+6]/x=(-3)[P(x)-P(x+4)Q_2(x)+6]/[m-P(x+4)+P(x)Q_1(x)]$
pick $n=\pm{3}$ by irreducible and improve above formula to :
$P(x)(Q_1(x)-1)+P(x+4)(Q_2(x)-1)=6-m,$$P(x)(Q_1(x)+1)-P(x+4)(Q_2(x)+1)=-6-m$
also note that:
$P(x)Q_1(x)-P(x)=P(x+4)-3x-m-P(x+4)Q_2(x)-nx+6,$$P(x)Q_1(x)+P(x)=P(x+4)-3x-m+P(x+4)Q_2(x)+nx-6$
then we can get:
$P(x+4)-3x-m-P(x+4)Q_2(x)-nx+6+P(x+4)(Q_2(x)-1)=6-m$
$P(x+4)-3x-m+P(x+4)Q_2(x)+nx-6-P(x+4)(Q_2(x)+1)=-6-m$
therefore, $m=\pm{6},n=\mp{3}$
