Prove/disprove the following: if $a^2$ divides $b^3$, then $a$ divides $b$
I've tried rewriting as $b^3=a^2c$, but I can't seem to show anything else for $c$. If I can prove that it is/isn't an integer, that would prove the whole thing for me.
$\begingroup$
$\endgroup$
6
-
4$\begingroup$ Counterexample: $a=2^3$, $b=2^2$. Then $a^2=2^6=b^3$, but obviously $a$ does not divide $b$, since it is smaller. $\endgroup$– CrostulSep 11, 2016 at 8:09
-
$\begingroup$ @Crostul Since $b$ is smaller, not $a$. $\endgroup$– user236182Sep 11, 2016 at 8:11
-
$\begingroup$ In general, you could pick $a = x^3$ and $b=x^2$ for some natural number $x$, then obviously $a^2 = x^6 = b^3$ but $a > b$ so $a$ cannot divide $b$. $\endgroup$– Cameron WilliamsSep 11, 2016 at 17:50
-
$\begingroup$ @CameronWilliams Where $x\ge 2$. $\endgroup$– user236182Sep 11, 2016 at 18:59
-
$\begingroup$ @user236182 right right. $\endgroup$– Cameron WilliamsSep 11, 2016 at 20:41
|
Show 1 more comment
2 Answers
$\begingroup$
$\endgroup$
If $a^2 \mid b^3$ then $a \mid b$.
How might one discover an answer? To disprove it, we'd need $a,b$ with $a \not \mid b$ but $a^2 \mid b^3$. Note that since $a^2 \mid b^3$, we must have every prime factor of $a$ appearing in $b$; so for $a \mid b$ to be false, we must have one prime factor of $a$ appearing to larger multiplicity in $a$ than in $b$. If we pick that multiplicity carefully, we might still be able to make sure that $a^2 \mid b^3$, though.
From here it's just a short leap to noting that if the prime factor has multiplicity $3$ in $a$ and $2$ in $b$, the conditions hold, and we have found a counterexample: $a = p^3, b = p^2$.
answered Sep 11, 2016 at 9:23
$\begingroup$
$\endgroup$
If $p$ is a prime, and $a=p^m$ and $b=p^n,$ with $m,n\in \mathbb N,$ then $a^2|b^3$ iff $2m\leq 3n$ iff $m\leq 3n/2,$ but $a|b$ iff $m\leq n.$
When $n\geq 2$ we have $3n/2=n+n/2\geq n+1.$
So if $n\geq 2$ and $a=p^{n+1}$ and $b=p^n$ we have $a>b$ but $a^2|b^3.$
answered Sep 11, 2016 at 17:46