6. can you find four examples that are true or a counterexample for the following statement? for every…

Explanation:

Step1: Recall the property of even and odd integers

Let (n) be an integer. An integer (n) can be either even ((n = 2k) for some integer (k)) or odd ((n=2k + 1) for some integer (k)).

Step2: Case 1: (n) is even

If (n = 2k), then (n^{2}+1=(2k)^{2}+1=4k^{2}+1). Since (4k^{2}) is even ((4k^{2}=2(2k^{2}))), (4k^{2}+1) is odd.

Step3: Case 2: (n) is odd

If (n = 2k + 1), then (n^{2}+1=(2k + 1)^{2}+1=4k^{2}+4k+1 + 1=4k^{2}+4k + 2=2(2k^{2}+2k + 1)). Since (2(2k^{2}+2k + 1)) is divisible by 2, it is even when (n) is odd. So a counter - example is (n = 1) (or any odd integer). When (n = 1), (n^{2}+1=1^{2}+1=2) (even).

Step4: Find four true examples

For (n = 0), (n^{2}+1=0^{2}+1 = 1) (odd). For (n = 2), (n^{2}+1=2^{2}+1=5) (odd). For (n = 4), (n^{2}+1=4^{2}+1 = 17) (odd). For (n=6), (n^{2}+1=6^{2}+1=37) (odd).

Answer:

Counter - example: (n = 1). True examples: (n = 0), (n = 2), (n = 4), (n = 6)