True or False. If the statement is true, give a proof. If it is false, give one example showing it is false. Suppose that a, b, c are integers.
a) If a|c and b|c then ab|c.
b) If a|bc then a|b or a|c.
c) If a|$b^2$ then $a^2$|$b^4$.
(a) is false; take a=4, b=6, and c=12.
(b) is also false; take a=8, b=6, and c=4.
Can someone help me out with part (c)?
On
Hint $\ $ It is a special case of the $ $ Divisibility Product Rule $ $ below
$\qquad \qquad\qquad \begin{align}&\ \ a\mid c\\ &\ \ \bar a\mid \bar c\\ \Rightarrow\ & a\bar a\mid c\bar c\\\end{align}\ \ $ by $ \begin{align} {an} &= c\\ \bar a \bar n &= \bar c\\\Rightarrow\ a\bar a n\bar n &= c\bar c\end{align}$
Remark $\ $ This is more familiar in fractional form
$\qquad\qquad\qquad\qquad \dfrac{c}a,\ \dfrac{\bar c}{\bar a}\in \Bbb Z\ \Rightarrow\ \dfrac{c\,\bar c}{a\,\bar a}\in\Bbb Z$
i.e. divisibility is closed under products boils down to $\,\Bbb Z\,$ being closed under products.
[Note: the exceptional cases $\,a,\bar a = 0\,$ need to be treated specially if you use fractions]
Hint: $a \mid b^2$ means that $b^2 = ka$ for some integer $k$. From here, can you find an integer $\ell$ such that $b^4 = \ell a^2$?