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.
 A: Write $t=x+3$. Now we have to prove that $Q=t^2-t+1$ divides $P=(t-1)^m+t^{2m+3}$. Note that if $a$ is root for $Q$ then we have $a^3=-1$. Note that $a-1 = a^2$. Now plug $a$ in to $P$ and we get:
\begin{eqnarray} (a-1)^m +a^{2m}\cdot a^3 &=& (a-1)^m -a^{2m} \\
&=& (a^2)^m-a^{2m} \\
&=& a^{2m}-a^{2m} \\ 
&=&0
\end{eqnarray}
So $a$ is also a root for $P$, and we can in the same manner deduce that the second root $b$ of $Q$ is also a root for $P$. Since $a\ne b$, it follows that $Q$ divides $P$. 
A: Note that 
$$(X+3)^3=X^3+9X^2+27X+27=(X^2+5X+7)(X+4)-1 $$
and
$$(X+3)^2 = X^2+6X+9=(X^2+5X+7)+(X+2).$$
A: Let us restate the value of $P$ first.
$P = (X+2)^m+(X+2)^{2m+3}+1^{2m+3} = (X+2)^m+(X+2)^{2m+3}+1 =$
$(X+2)^m+(X+2)^{2m+3}+(X+2)^0 = (X+2)^{3m+3} = (X+2)^{(m+1)^{3}}$
And, Let us restate the value of $Q$.
$Q = X^2+4X+4+X+3 = X^2+4X+4+X+2+1 = (X+2)^2+(X+2)^1+(X+2)^0$
Do you notice the relationship between $P$ and $Q$?
A: We can also prove it by induction.
$P_0(x)=1+(x+3)^3=x^3+9x^2+27x+28=(x^2+5x+7)(x+4)\quad\checkmark$
$\begin{align}
P_{m+1}(x) &=(x+2)^{m+1}+(x+3)^{2m+5}\\
&=(x+2)^m(x+2)+(x+3)^2\overbrace{\big((x^2+5x+7)Q_m(x)-(x+2)^m\big)}^{\text{induction hypothesis}}\\\\
&=(x+2)^m\underbrace{(x+2-x^2-6x-9)}_{-x^2-5x-7}+(x+3)^2(x^2+5x+7)Q_m(x)\\\\
&= (x^2+5x+7)Q_{m+1}(x)\quad\checkmark
\end{align}$
