True or False: If $a$|$b^2$ then $a^2$|$b^4$.

Question

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)?

It is likely easier to see that "if $c\mid d$ then $c^2 \mid d^2$." Then apply this with $c=a$ and $d= b^2$ — quid, Sep 01 '16 at 21:30

score 4 · Answer 1 · answered Sep 01 '16 at 21:25

4

Hint: $a \mid b^2$ means that $b^2 = ka$ for some integer $k$. From here, can you find an integer $\ell$ such that $b^4 = \ell a^2$?

answered Sep 01 '16 at 21:25

user133281

16,073

$(k^2a^2)=la^2$ ==> $l=k^2$ – BrianW Sep 01 '16 at 21:29
And since k is an integer, then l is also an integer because Z is closed under multiplication. Does that satisfy as a proof? – BrianW Sep 01 '16 at 21:35

Bill Dubuque · Answer 2 · 2016-09-01T23:50:59.233

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]

True or False: If $a$|$b^2$ then $a^2$|$b^4$.

2 Answers2