Show that $Q=X^2+5X+7$ divides $P=(X+2)^m+(X+3)^{2m+3}$ for any $m\in\Bbb N$ Let $$P=(X+2)^m+(X+3)^{2m+3}$$ and $$Q=X^2+5X+7.$$ I need to show that $Q$ divides $P$ for any $m$ natural. 
I said like this: let $a$ be a root of $X^2+5X+7=0$. Then $a^2+5a+7=0$. 
Now, I know I need to show that $P(a)=0$, but I do not know if it is the right path since I have not found any way to do it.

Note that 
$$(X+3)^3=X^3+9X^2+27X+27=(X^2+5X+7)(X+4)-1 $$
and
$$(X+3)^2 = X^2+6X+9=(X^2+5X+7)+(X+2).$$
