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.
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Sign up to join this communityProve/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.
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: $a = p^3, b = p^2$.
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 $a^2|b^3.$