Show that $Q=X^2+5X+7$ divides $P=(X+2)^m+(X+3)^{2m+3}$ for any $m\in\Bbb N$

Question

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.

Answer 1

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 $P$.

Answer 2

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

Answer 3

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 $Q$?

Answer 4

We can also prove it by induction.

$P_0(x)=1+(x+3)^3=x^3+9x^2+27x+28=(x^2+5x+7)(x+4)\quad\checkmark$

$\begin{align} P_{m+1}(x) &=(x+2)^{m+1}+(x+3)^{2m+5}\\ &=(x+2)^m(x+2)+(x+3)^2\overbrace{\big((x^2+5x+7)Q_m(x)-(x+2)^m\big)}^{\text{induction hypothesis}}\\\\ &=(x+2)^m\underbrace{(x+2-x^2-6x-9)}_{-x^2-5x-7}+(x+3)^2(x^2+5x+7)Q_m(x)\\\\ &= (x^2+5x+7)Q_{m+1}(x)\quad\checkmark \end{align}$