Question

Prove that for an integer k, k² + 4k + 6 is odd if and only if k is odd.

Answer 1

To prove:

For an integer k:

                        (1)

is odd if and only if k is odd

Proof by Contradiction:
Step 1:

By Contradiction:

Assume k is even.

Step 2:
Then, by theorem, as the square of even integer is even, k² is even.

Step 3:

By theorem, 4k, the product of an even integer by 4 is even.

Step 4:

6 is an even integer.

Step 5:
By theorem, the sum of 3 even integers is even.

Thus,

is even.

Step 6:
But, Given:

is odd.

This contradiction proves that the initial assumption that k is even is incorrect.

This proves the required result that

is odd if and only if k is odd.

Also:
If k is odd, since by theorem, the square of an odd integer is odd, k² is odd.

4k = the product of an even integer by 4 is even.

6 is even integer.

Thus,

is the sum of an odd integer, an even integer and an even integer.

Thus, by Theorem,

is an odd integer.