Question

How to prove this ? `2|k^2` then `2|k` for some `k\inZZ`

Answer 1

See below.

Explanation:

Proof 1: Suppose that `2 | k^2`, but `2 \cancel(|) k`. This means that `k^2` is even, and `k` is odd. But the square of ad odd number is still odd, since

`(2k+1)^2 = 4k^2+4k+1`

so, it is impossible that `k` is odd and `k^2` is even.

Proof 2: If a number is a perfect square, all the exponents in its prime factorization must be even. Since `k^2` is a perfect square, its prime factorization must be like

`k^2 = 2^m * r`

where `m` is even and `r` is the rest of the factorization. In fact, `2` divides `k^2` by hypothesis, and `m` must be even because `k^2` is a square. Since `k=sqrt(k^2)`, the factorization of `k` will contain `2^{m/2}`, which means that `2` divides `k`.

Again, we're shown that it is impossible that `2` divides `k^2` without dividing `k`.