prove problem if $m^2 | n^2, \text{then } m|n$

Question

Prove that if $m^2 | n^2, \text{then } m|n$
if $m^2 | n^3, \text{then } m|n$? prove or disprove.

i really no idea of that, someone could help me?

Do you know the fundamental theorem of arithmetic, that every integer $n>1$ has a unique factorization as a product of primes? — Brian M. Scott, Feb 18 '13 at 08:56
@ulead86: Indeed it is, and that question has a couple of excellent answers, one using the fundamental theorem and the other avoiding it. — Brian M. Scott, Feb 18 '13 at 08:59
Or as addition to Andrew $m=a^3$, $n=a^2$ is for every $a>1$ a counterexample — Dominic Michaelis, Feb 18 '13 at 09:15

score 1 · Accepted Answer · answered Feb 18 '13 at 09:29

For 1. there are a lot of proofs. This one is my Favorite $$m^2|n^2 \implies k = \frac{m^2}{n^2} = \left(\frac{m}{n}\right)^2 $$ But because only Integers have Integer roots, $\frac{m}{n}$ must be an Integer.

For 2. Let $a$ be an Integer greater than 1. set $m=a^3$ and $n=a^2$. We see $m^2 = a^6 = n^3$ so $m^3| n^2$ but $a^3$ is not a divisor of $a^2.$

prove problem if $m^2 | n^2, \text{then } m|n$

1 Answers1