Inductive proof of 2|($n^2$ +3n + 2) if n is a natural number

Question

I was looking at an example problem:

Please prove the following statement: if n is a natural number then $\displaystyle2|( n^2 + 3n + 2)$

In the example solution it showed:

Proof: P(n) be $\displaystyle2|(n^2 + 3n + 2)$

Bases Step: let n = 1 so: $2|(1^2 + 3(1) + 2)$ ->$ 2|6$, thus it holds for n

Induction Hypothesis: for some $n \geq 1, 2|( n^2 + 3n + 2)$

Induction Step:

$2|( n^2 + 3n + 2)$ by hypothesis and $2|2(n+2)$

->$2|(n^2 + 2n + 1 + 3n + 2)$

->$2|[(n + 1)^2 + 2(n+1)+ 2]$

proving that:

$\displaystyle2|( n^2 + 3n + 2)$

I'm confused at how:

$\displaystyle2|( n^2 + 3n + 2)$

got to:

$2|(n^2 + 2n + 1 + 3n + 2)$

Also how come we are using $2|2(n+2)$ can't we just use $2|2n$ or $2|2(n+1)$

Answer 1

I think that after you realize that $2|(n^2+3n+2)$ and $2|2(n+2)$, you should get that $$ 2|[n^2+3n+2+2(n+2)] $$ which is equivalent to $$ 2|[n^2+2n+1+3(n+1)+2] $$ which is equivalent to $$ 2|[(n+1)^2+3(n+1)+2] $$ In the end you wanted this last identity to complete the induction, not $$ 2|[(n+1)^2+2(n+1)+2] $$ as you wrote.

Answer 2

We know that, if $a|b$ and $a|c$,then $a|(b+c)$ which is obvious. And our aim is to prove that $P(n+1)$ statement is true. So, we have to bring our $P(n+1)$ expression : $(n+1)^2 + 3(n+1) + 2$. from our P(n) expression: $n^2+3n+2$. So, we add $2(n+2)$ and not $2n$ or $2(n+1)$

Answer 3

We need to show that if $n^2+3n+2$ is even then $(n+1)^2+3(n+1)+2$ is even. The latter expression is $n^2+2n+1+3n+3+2$. Reorder the terms and you get $$n^2+3n+2+(2n+1+3)$$ It's easy to see that $2n+1+3$ is even, since $2n+1+3=2n+4=2(n+2)$. The sum of two even numbers is even, and so we're done.

There seems to be a typo in what you wrote. You say that you want to show $$2|(n^2 + 2n + 1 + 3n + 2)$$ but what you really want to show in the induction step is that $2$ divides $n^2+2n+1+3n+3+2$, i.e., your expression was missing a $3$.

Answer 4

The solution is making the problem more difficult than it needs to be (as is often the case in "prove this result by induction" problems).

Instead, observe that $$n^2+3n+2=n^2+2n+n+2=n(n+2)+n+2=(n+1)(n+2).$$ Now, suppose $n$ is a natural number. If $2\mid n,$ then $2\mid(n+2),$ and so $2\mid(n+1)(n+2),$ and we're done. Otherwise (if $2\not\mid n$), we have $2\mid(n+1),$ so $2\mid(n+1)(n+2),$ and we're done.

Answer 5

If a proof without induction is allowed; we split it into two cases:

Case 1: if n is even/zero, proof is trivial.

Case 2: if n is odd, let $n=2k-1$. Now, $n^2+3n^2+2=n^2+3n^2=4k^2+2k-2=0\mod 2$

It can be used for final step of induction as well.

Answer 6

We can factorize it as $(n+2)(n+1)$ so one of them will be even when n is a natural number i hope you know why. Its simple observation. Thus $2|n^2+3n+2$ thats all.