Suppose the integers $m^2$ and $n$ are relatively prime. Show that $m$ and $n^2$ are relatively prime. My attempt so far:
Since $m^2$ and $n$ are relatively prime, $am^2 + bn = 1$ for some integers $a$ and $b$. 
I know that I will have to use this to somehow prove that $cm + dn^2 = 1$ for some integers $c$ and $d$, but I'm not sure how to do this. 
 A: Start with $$am^2 + bn = 1$$ and multiply by $bn$ to get $$abm^2n + b^2n^2 = bn.$$ Since $bn = 1 - am^2$ this can be written as $$abm^2 n + b^2n^2 = 1 - am^2$$ which may in turn be written as $$abm^2n + am^2 + b^2n^2 = 1.$$ Thus $$(abmn + am)m + (b^2)n^2 = 1$$ so you can take $c = abmn + am$ and $d = b^2$.
A: Good start. Now note that 
$$1=(am^2+bn)^2=(a^2m^3+2ambn)m+(b^2)n^2.$$
A: If $p$ a prime divides both $m$ and $n^2$, then $p$ divides $n$ by primality, hence $p$ divides $m^2$ and $n$, contradicting the fact that $\gcd(m^2, n) = 1$. Thus, $m$ and $n^2$ are relatively prime.
A: All one-letter variables are integers.
(i). If $1<u\in N$ then u is divisible by a prime. Proof: Let $v$ be the least $w$ such that $1<w\leq u$ and $w|u.$ Then $v$ is prime, for if $ v=w_1 w_2$ with $w_1>1<w_2 ,$ then  $w_1$ is a divisor of $u$ with $1<w_1<w,$ contrary to the def'n of $w.$
(ii). If $p$ is prime and $p|n^2$ then $p|n$. Proof: If not then $\gcd(p,m)=1$ because $p$ is prime, implying there are $a,b$ with $1=a p+b n.$ But then $$n= n a p+b n^2=(n a +b (n^2/p))p$$ and $n^2/p$ is an integer, so $k=n a+b(n^2/p)$ is an integer, and $n= k p,$ so $p|n.$  
(iii). If  $\gcd (m,n^2)> 1$ then by (i), some prime $p$ divides both $m$ and $n^2.$ But then by (ii), $p$ divides $n,$ so p divides both $m^2$ and $n,$ implying $\gcd (m^2,n)\geq p>1.$
