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

Jun 20, 2018

See below.

Explanation:

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

${\left(2 k + 1\right)}^{2} = 4 {k}^{2} + 4 k + 1$

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

${k}^{2} = {2}^{m} \cdot r$

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 ${2}^{\frac{m}{2}}$, which means that $2$ divides $k$.

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