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Prove or disprove: $\forall a,b\in \mathbb N $, $a$ and $b$ are coprime integers IFF $a^2$ and $b^2$ are coprime integers.

I know how to prove that if $a$,$b$ are coprime integers but I have no clue how to prove the other way around to satisfy the IFF.

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1  
Hint: $p|a^2$ implies $p|a$. –  mathemagician Jan 23 '13 at 18:55
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Show what you've done so far, that way people have something to work with. –  Erik G. Jan 23 '13 at 18:55
    
I'm having a hard time translating it to English –  Georgey Jan 23 '13 at 19:19

3 Answers 3

up vote 4 down vote accepted

How to argue depends on what you know about coprime integers.

I.

For example, if you know that $a,b$ being coprime means that there are integers $n,m$ with $$ an+bm=1, $$ you can argue as follows:

First, if $a^2$ and $b^2$ are coprime, as witnessed by integers $n,m$, so $$ a^2n+b^2m=1, $$ then $k=an$ and $l=bm$ witness that $a,b$ are coprime: $$ ak+bl=1. $$

The other direction is more interesting: First, note that if $x,y$ are coprime, so are $x^2,y$: $$ xn+ym=1\Longrightarrow (xn+ym)^2=1, $$ but $$ (xn+ym)^2= x^2 n^2+y(2xnm+ym^2)= x^2k+yl, $$ with $k=n^2$ and $l=2xnm+ym^2$.

This gives us that if $a,b$ are coprime, $a^2,b$ are coprime. But then, letting $x=b,y=a^2$, this gives us that $x^2=b^2$ and $y=a^2$ are coprime as well, as we wanted.

II.

On the other hand, if you know that being coprime can be verified in terms of common prime divisors, then the argument is simpler, because if $p$ is prime and $p|xy$ then $p|x$ or $p|y$. This means that for any prime $p$, we have that $p|a^2$ iff $p|a$, and the same with $b^2$ and $b$, so the prime divisors of $a^2,b^2$ are the same as those of $a,b$. Hence, one pair is coprime iff the other is coprime.

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the definition of coprime integers I work with says that there's no $k\in \mathbb N >1$ that $k|a$ and $k|b$, that's it. –  Georgey Jan 23 '13 at 19:44
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The Bezout two-step argument can be replaced by $1=(xn+ym)^3=x^3n^3+3x^2n^2ym+3xny^2m^2+y^3m^3=n^2(X)+m^2(Y)$. –  André Nicolas Jan 23 '13 at 20:27

let $d|a$ and $d|b$ then $d^2|a^2$ and $d^2|b^2$. So - by considering the contrapositive - if $a^2$ and $b^2$ are coprime, $a$ and $b$ are.

If $a$ and $b$ are coprime then $a^2$ and $b^2$ must be too, if a prime $p|a^2$ then $p|a$ so if a prime divided $a^2$ and $b^2$ it would be a common factor of $a$ and $b$.

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Hint $\rm\ (a,b)^2 = (a^2,b^2)\,$ by cancelling $\rm\,(a,b)\,$ in $\rm\ (a,b)^3 = (a^3, a^2 b, ab^2, b^3) = (a,b) (a^2, b^2).\:$
Hence $\rm\,(a,b)=1\iff (a^2,b^2) = 1.$

Remark $\ $ The proof utilizes only basic gcd laws (associative, commutative, distributive, etc), e.g. see here for background. It easily generalizes to the Freshman's Dream $\rm\: (a,b)^n = (a^n,b^n).\:$

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