To prove that if 7n4 is even, then n must be even, we’ll start by analyzing the properties of even and odd numbers.
1. **Understanding Even and Odd Numbers**: A number is even if it is divisible by 2. Conversely, a number is odd if it is not divisible by 2. We know that:
- If n is even, then n = 2k, where k is an integer.
- If n is odd, then n = 2k + 1, where k is an integer.
2. **Analyzing the Expression 7n4**: The expression 7n4 can be examined under both conditions (i.e., when n is even and when n is odd).
3. **Case 1: n is even**
- If n is even, then n4 = (2k)4 = 16k4, which is clearly an even number.
- Thus, multiplying by 7, 7n4 = 7 imes 16k4 = 112k4, which is also even.
4. **Case 2: n is odd**
- Now consider when n is odd, so let n = 2k + 1.
- This gives us: n4 = (2k + 1)4. The binomial expansion of (2k + 1)4 results in a polynomial that is clearly odd as it includes all terms with an even exponent multiplied by one (even) and also has a constant term on expansion which is odd.
- When we multiply this by 7, the product 7n4 remains odd since the product of an odd number (7) and an odd number (n4) results in an odd number.
5. **Conclusion**: Since we have shown:
- If n is even, then 7n4 is even.
- If n is odd, then 7n4 is odd.
That implies if 7n4 is even, then n must be even as well. This proof confirms the statement, completing our proof.