Answer to Question #248542 in Discrete Mathematics for senpaisuz

Question

Prove that for any integer n

n, if n

n is an odd integer, then 6n

2

+5n+1

6n2+5n+1 is an even integer.

Answer 1

Suppose n is odd. Then, $n=2k-1, k\in \N$

$6n^2+5n+1=6(2k-1)^2+5(2k-1)+1\\ =6(4k^2-4k+1)+10k-5+1\\ =24k^2-24k+6+10k-4\\ =24k^2-14k+2\\ =2(12k^2-7k+1)\\ =2M, M=12k^2-7k+1\\ \text{Hence } 6n^2+5n+1 \text{ is even}$