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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.

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  • $\begingroup$ Have you thought about the factorizations of $m$ and $n$ into primes? $\endgroup$ – Catalin Zara Jun 16 '16 at 2:33
  • $\begingroup$ No I hadn't thought of that... but how could I do that if I don't know the values of $m$ and $n$? Would I use the Euclidean Algorithm somehow? $\endgroup$ – maths123 Jun 16 '16 at 2:37
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    $\begingroup$ From the definition of primality, a prime $p$ with $p|n^2$ has $p|n$. $\endgroup$ – anomaly Jun 16 '16 at 2:50
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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 $d = b^2$.

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  • $\begingroup$ Thanks! In an exam would I have to further prove that c and d are integers, or could I just leave it at that? $\endgroup$ – maths123 Jun 16 '16 at 2:54
  • $\begingroup$ They are integers since they are basic expressions of other integers. (addition and multiplication) $\endgroup$ – Jack Tiger Lam Jun 16 '16 at 3:24
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Good start. Now note that $$1=(am^2+bn)^2=(a^2m^3+2ambn)m+(b^2)n^2.$$

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  • $\begingroup$ Thanks! So would it be enough to say that $1 = am^2 + bn$ so $1^2 = (am^2 + bn)^2 $$\iff 1 = a^2m^4 + b^2n^2 + 2am^2bn \iff 1 = (a^2m^3 + 2abnm)m + b^2(n^2)$ and since $(a^2m^3 + 2abnm)$ and $b^2$ are the result of addition and multiplication of integers then they will also be integers? $\endgroup$ – maths123 Jun 16 '16 at 2:51
  • $\begingroup$ If you want to add anything, say that there is an integer linear combination of $m$ and $n^2$ which is equal to $1$, so $m$ and $n^2$ are relatively prime. I think there is no need to prove that $a^2m^3+2abmn$ and $b^2$ are integers. Note that I prefer the approach of Sameer Kailasa, but I wanted to complete the argument along the lines you had started. $\endgroup$ – André Nicolas Jun 16 '16 at 2:56
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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.

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  • $\begingroup$ Thanks, I wouldn't have thought of doing it this way :) $\endgroup$ – maths123 Jun 16 '16 at 2:54
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    $\begingroup$ Thanks @Sameer! This was very helpful! +1 $\endgroup$ – user345872 Jul 1 '16 at 18:43
  • $\begingroup$ Haha, go Huskies! $\endgroup$ – Sameer Kailasa Jul 1 '16 at 21:01
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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 $\gcd (m^2,n)\geq p>1.$

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  • $\begingroup$ What a wonderful proof, thank you :) $\endgroup$ – maths123 Jun 16 '16 at 22:10

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