<s>[INST] 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.
 [/INST] We can also prove it by induction.
$P_0(x)=1+(x+3)^3=x^3+9x^2+27x+28=(x^2+5x+7)(x+4)\quad\checkmark$
$\begin{align}
P_{m+1}(x) &=(x+2)^{m+1}+(x+3)^{2m+5}\\
&=(x+2)^m(x+2)+(x+3)^2\overbrace{\big((x^2+5x+7)Q_m(x)-(x+2)^m\big)}^{\text{induction hypothesis}}\\\\
&=(x+2)^m\underbrace{(x+2-x^2-6x-9)}_{-x^2-5x-7}+(x+3)^2(x^2+5x+7)Q_m(x)\\\\
&= (x^2+5x+7)Q_{m+1}(x)\quad\checkmark
\end{align}$
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