Prove n even iff 7n+4 even

To prove. For a positive integer n: n is even ⇔ 7n + 4 is even. We prove both directions of the biconditional.

Part 1: if n is even, then 7n + 4 is even (direct proof)

1. Let n be even, so n = 2k for some integer k.
2. Then 7n + 4 = 7(2k) + 4 = 14k + 4 = 2(7k + 2).
3. 7k + 2 is an integer, so 7n + 4 is 2 times an integer. Hence 7n + 4 is even.

Part 2: if 7n + 4 is even, then n is even (proof by contraposition)

We prove the contrapositive: if n is odd, then 7n + 4 is odd.

1. Let n be odd, so n = 2k + 1 for some integer k.
2. Then 7n + 4 = 7(2k + 1) + 4 = 14k + 11 = 2(7k + 5) + 1.
3. This has the form 2m + 1, so 7n + 4 is odd.

The contrapositive is true, so the original statement is true: if 7n + 4 is even, then n is even.

Conclusion

Both implications hold, so n is even if and only if 7n + 4 is even. ∎