I have proven that $a^5|b^3$ implies that $a\mid b$ using prime factorization and showing that every prime factor of $a$ will divide every prime factor of $b$. I have to prove or disprove that the hypothesis also implies that $a^2|b$, however, I am not sure whether I should try to prove or disprove this. I think if I knew which way to work towards, I could figure it out, but I'm not sure where to start.
If $a^5|b^3$, does $a^2|b$?
1
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number-theory
elementary-number-theory
divisibility
prime-factorization
3 Answers
3
No: if $a=8$ and $b=32$, then $a^5=2^{15}=32^3$, but $a^2=64$ does not divide $b$.
3
Let $a=p^\alpha,b=p^\beta$.
We should have
$$5\alpha\le3\beta\implies2\alpha\le\beta,$$ which does not hold. (Smallest counterexample: $3,5$, with $p=2$).
2
use $a=2^3$ and $b=2^5$ and you have a counterexample.