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

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

using mathematical induction...
help please

thank you

2. 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)

3. Hello, NeedHelp18!

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

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

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

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

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

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

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

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

4. $n^{2}+n+2=\frac{4n^{2}+4n+8}{4}=\frac{(2n+1)^{2}+7 }{4}.$

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