Show that $x^2+9x+20$ is divisible by 2 for all $x \in \mathbb{Z}$ I'm having extremely hard time getting how proof by induction should work for this case.
This is my attempt so far:
(1) When $x = 1$
$1^2 + 9 + 20 = 30$ which is divisible by 2.
(2) Now assume that it is true when $x = k$
$k^2 + 9k + 20$ is divisible by 2
(3) Then $x = k + 1$
$(k + 1)^2 + 9(k + 1) + 20 = k^2 + 2k + 1 + 9k + 9 + 20 = k^2 + 11k + 30$
And now I'm stuck. I could continue it with $k(k + 11) + 30$ but I don't understand how would that bring me closer to the solution. Maybe my approach is wrong? Any help would be appreciated, thank you.
 A: Because  $$x^2+9x+20=(x+4)(x+5).$$
A: We have $$x^2+9x+20$$
We can factorise this to $$(x+4)(x+5)$$
Take $r=x+4$. We now have $$r(r+1)$$For some integer input $x$, and as a result, a integer input $r$, we have an expression for the product of two consecutive integers. One of these must be odd and the other must be even.  We know that an even integer multiplied by an odd integer must be even, and so the output of your expression must be even too.
A: Write the given expression as $x^2+x+8x+20$; the last two terms are individually even always.
The first two terms can be written as $x(x+1)$, a product of two consecutive integers. One of them is always even so the product too is even.
A: We have 
$$\begin{align}
\text{odd}\times \text{odd}&=\text{odd}, \\
\text{odd}\times \text{even}&=\text{even}, \quad \text{and}\\
\text{even}\times \text{even}&=\text{even}
\end{align}$$ and
$$\begin{align}
\text{odd}+ \text{odd}&=\text{even}, \\
\text{odd}+ \text{even}&=\text{odd}, \quad \text{and}\\
\text{even}+ \text{even}&=\text{even}.
\end{align}$$ 
Thus we have, when $x$ is odd, that
$$\begin{align}
x^2+9x+20&=\text{odd}^2+\text{odd}(\text{odd})+\text{even} \\
&=\text{odd}+\text{odd}+\text{even}\\
&=\text{even}+\text{even}\\
&=\text{even}.
\end{align}$$
This is equivalent to using modular arithmetic.
Can you do the case when $x$ is even?
A: We want to show that $(k+1)^2+9(k+1)+20$ is divisible by $2$ given that $k^2+9k+20$ is divisible by $2$.
$[(k+1)^2+9(k+1)+20]-(k^2+9k+20)=(2k+1)+9=2(k+5)$ which is divisible by $2$.
Hence $[(k+1)^2+9(k+1)+20]$ is divisible by $2$.
Also, as mentioned by Cameron, we have to consider non-positive numbers as well.
A: $$
x^2+9x+20 = (x+4)(x+5)
$$
so for any $x \in \mathbb{Z}$ is an even number.
A: First method:
$x=2k\Rightarrow x^2+9x+20=4k^2+18k+20=2(2k^2+9k+10)$
while
$x=2k+1\Rightarrow 
\color{red}{x^2}+\color{green}{9x}+\color{blue}{20}=
\color{red}{4k^2+4k+1}+\color{green}{18k+9}+\color{blue}{20}=
2(2k^2+11k+15)$
Second method (by induction):
Let $p(x)=x^2+9x+20$ and let $P(x)$ be $2\vert p(x) \forall x\in\mathbb Z$, to be proved.
Then $p(0)=20$ and $p(1)=30$ so $P(0)$ and $P(1)$ are true.
Suppose $P(x)$ is true for some $x>0$, this is $2\vert p(x)$. Then $2\vert p(x+1)$ if and only if $2\vert p(x+1)-p(x)$.
$p(x+1)-p(x)=(x+1)^2+9(x+1)+20-(x^2+9x+20)=2(x+5)$
so $2\vert p(x+1)-p(x)$ and $P(x+1)$ is also true.
By the way, $2\vert p(x)-p(-x)=18x$ so $P(x)$ is true for all $x\in\mathbb Z^-\cup \left\{0\right\}\cup\mathbb Z^+$.
A: For fun;
$x(x+9)+20 =$
$(x+10-10)(x+9) +20=$
$(x+10)(x+9) -10(x+9) +20$.
1) $2$ divides $(x+10)(x+9)$ (Why?)
2) $2$ divides $(-10(x+9))$ (Why?)
3) $2$ divides $20$ ,
hence $2$ divides the sum of the above terms.
