Prove that $2|(x^4-3) <=> 4|(x^2+3)$

Question

Prove that $2|(x^4-3) <=> 4|(x^2+3)$

What i have right now is: Consider the case (=>): Since $x^4-3$ divides $2$ then, there must exist n belongs to integer, such that $n = \frac{x^4-3}{2}$

I have no idea what should I do next. Any helpful hints?

Answer 1

The details depend on how formal you want to be.

Suppose that $2$ divides $x^4-3$. Then $x$ is odd. Thus $x=2y+1$ for some integer $y$. Then $x^2=4(y^2+y)+1$, and therefore $4$ divides $x^2+3$.

The other direction is easier.

Answer 2

$$2\mid (x^4-3)\implies\;x\;\;\text{is odd (why?)}\;\implies x=\pm1\pmod 4\implies x^2=1\pmod 4\implies$$

$$x^2+3=0\pmod 4$$

Try now the other direction working your way modulo $\;4\;$ as above (in fact, only for the very beginning you'd need that) .

Answer 3

$$2|(x^4-3) \iff 2|(x^4-3-6) \iff 2|(x^2-3)(x^2+3)$$

Since $2$ is prime it must divide one of those factors. And

$$2|(x^2+3)\iff 2|(x^2+3-6) \iff 2|(x^2-3)$$

So $2$ divides $both$ factors if it divides either one!

Answer 4

Simpler solution: $$\hbox{LHS}\quad\Leftrightarrow\quad\hbox{$x$ is odd}\quad\Leftrightarrow\quad\hbox{RHS}\ .$$ The first $\Leftrightarrow$ should be obvious, so should the second $\Leftarrow\,$. The second $\Rightarrow$ takes a small amount of work.