If $4\mid a+bc$ and $6\mid b+ac$ prove that $2\mid a^2-b^2$

If $4\mid a+bc$ and $6\mid b+ac$ prove that $2\mid a^2-b^2$

This is as far as I get:

$$a + bc = 4k\qquad\text{for some $k\in\Bbb Z$} \\b+ac =6l\qquad\text{for some $l\in\Bbb Z$}\\\implies (a^2-b^2)(1-c^2) = 16k^2 + 36l^2$$

Solution 1:

Following your approach, $$a + bc = 4k \implies a^2 + abc = 4ak,\tag 1$$ and $$b+ac =6l \implies b^2 + abc = 6bl. \tag 2$$ Subtract $(2)$ from $(1)$ to arrive at $$a^2-b^2 = 4ak-6bl = 2(2ak-3bl).$$ Q.E.D.

Solution 2:

Sometimes it pays off to think in much simpler terms, just involving parity. If $a$ is even, then so is $b$, because $b + ac$ is even. If $a$ is odd, then so are both $b$ and $c$, because $a + bc$ is even.

So $a$ and $b$ have the same parity, from which the result follows.