# How to prove that n^2 + n + 2 is even??

• Nov 30th 2008, 11:27 AM
NeedHelp18
How to prove that n^2 + n + 2 is even??
using mathematical induction...

thank you
• Nov 30th 2008, 11:38 AM
running-gag
It's a pity to use induction
n²+n+2 = (n+1)² - n + 1
n odd => (n+1)² even => (n+1)² - n + 1 even
n even => (n+1)² odd => (n+1)² - n + 1 even

By induction
Easy for n=0
If right for n then n²+n+2 even
(n+1)²+(n+1)+2 = n²+2n+1 + n + 3 = n²+n+2 +2n+2= n²+n+2 +2(n+1)
• Nov 30th 2008, 12:22 PM
Soroban
Hello, NeedHelp18!

Quote:

Prove by induction: .$\displaystyle n^2 + n + 2$ is even.

Verify $\displaystyle S(1)\!:;\;1^2 + 1 + 2 :=\:4$ . . . even!

Assume $\displaystyle S(k)\!:\;\;k^2 + k + 2 \:=\:2a\;\text{ for some integer }a$

Add $\displaystyle 2k+2$ to both sides: .$\displaystyle (k^2+k+2) + (2k+2) \;=\;2a + (2k+2)$

We have: .$\displaystyle k^2+3k + 4 \;=\;2a + 2k+2$

. . $\displaystyle (k^2 + 2k + 1) + (k + 1) + 2 \;=\;2a + 2k + 2$

. . $\displaystyle \underbrace{(k+1)^2 + (k+1) + 2}_{S(k+1)} \;=\;\underbrace{2(a + k + 1)}_{\text{even}}$

And we have proved $\displaystyle S(k+1)$ . . . The inductive proof is complete.

• Nov 30th 2008, 01:24 PM
Krizalid
$\displaystyle n^{2}+n+2=\frac{4n^{2}+4n+8}{4}=\frac{(2n+1)^{2}+7 }{4}.$

$\displaystyle 2n+1$ is odd hence $\displaystyle (2n+1)^2$ is odd and $\displaystyle (2n+1)^{2}+7$ is even, therefore $\displaystyle n^{2}+n+2$ is even.