Which of the following is the correct proof to show that the square of an even integer is an even integer?
1) Assume that an integer n is odd. By definition of an odd integer, n = 2k + 1.
n² = (2k + 1)² = 4k² + 4k + 1 = 2 (2k² + 2k) + 1.
The square of an odd integer is a multiple of 2, therefore it is even. By contradiction, the square of an even integer is an even integer.
2) 4² = 16
3) An even integer is a multiple of 2. 2² = 4. 4 is a multiple of 2; therefore, the square of an even integer is even.
4) An even integer n = 2k where k is an integer by definition of an even integer. n² = (2k)² = 4k² = 2 (2k²). The square of an even integer is a multiple of 2, therefore it is even.