Thread: Test Review

I am to prove this statement:

For all integers n, n^2-n+3 is odd.

This is what I did, I just want to make sure its logical.

Suppose n is an integer such that n^2-n+3 is odd. this means there exists an integer, a, such that 2a+1 = n^2-n+3. It then follows that 2a = n^2-n+2 which is, by definition even. Since n^2-n+2 and n^2-n+3 are consecutive integers then it follows that n^2-n+3 is odd.

Originally Posted by algebrapro18

I am to prove this statement:

For all integers n, n^2-n+3 is odd.

The integer n is either even or odd. If it is even then n^2 - n +3 is odd. If it is odd then n^2 - n + 3 is odd. Thus, n^2 - n + 3 is always odd.

So it would be something like this?

Case 1: Let integer n be even. That means there exists some integer c such that 2c = n. n^2-n+3 = (2c)^2-2c+3 = 4c^2-2c+3 = 2(2c^2-c). Since 2c^2-c is an integer then n^2-n+3 is odd by definition.

Case 2: Let integer n be odd. That means there exists some integer c such that 2c+1 = n. n^2-n+3 = (2c+1)^2-(2c+1) +3 = 4c^2+4c+1-2c-1+3 = 4c^2+2c+3 = 2(2c^2+c) + 3. Since 2c^2+c is an integer then n^2-n+3 is odd by definition.