Prove the following statement: The sum of any two consecutive integers can be written in the form 4n + 1 for some integer n.

What will be an ideal response?

Proof: Suppose n is any integer. By the parity principle, either n is even or n is odd, so we consider
two cases.
Case 1 (n is even): By definition of even, there is an integer a such that n = 2a. The next
consecutive integer after n is n + 1, and so
n + (n + 1) = 2a + (2a + 1) = 4a + 1 = 2(2a) + 1:
Now 2a is an integer because products of integers are integers. Thus n + (n + 1) equals 2 times an
integer plus 1, and so n + (n + 1) is odd by definition of odd.
Case 2 (n is odd): By definition of odd, there is an integer a such that n = 2a + 1. Then
n + (n + 1) = (2a + 1) + [(2a + 1) + 1] = 4a + 2 + 1 = 2(2a + 1) + 1:
Now 2a + 1 is an integer because products and sums of integers are integers. Thus n + (n + 1) equals
2 times an integer plus 1, and so n + (n + 1) is odd by definition of odd.
Conclusion: In both possible cases n + (n + 1) is odd.
Prove the following statement: The sum of any two consecutive integers can be written in the form
4n + 1 for some integer n.

Mathematics