<s>[INST] Let $a$, $b$ and $c$ be integers. Prove that if $4|(a+bc)$ and $6|(b+ac)$, then $2|(a^2-b^2)$. This is my work so far.

How can I isolate the $(a^2-b^2)$?
I am not asking for a full solution, just a hint and next step.

All done! Thanks Michael.

 [/INST] $4|(a+bc)$ then $a+b c = 4k\to c=\dfrac{4 k-a}{b}$
$6|(b+ac)$ then $b+ac=6h$ plug $c$
$b + a \,\dfrac{4 k-a}{b}=6h$
$b^2+4ak-a^2=6bh$
$a^2-b^2=2 (2 a k-3 b h)$
Which means that $2|(a^2-b^2)$
QED
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