I have a question from a sample exam I find difficults to solve:
Prove that if $\gcd(a, b) = 1 ⇒ \gcd(a^2, b^2) = 1$ .
I don't have any idea how to start. I'd like to get helped. thanks!
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What setting? Ordinary integers ($\mathbb{Z}$), a UFD, a general ring? – Daniel Fischer Jul 08 '13 at 12:49
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I'm sure this has been asked and answered here before, but how to find it? – Gerry Myerson Jul 08 '13 at 12:50
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And if over $\mathbb{Z}$, what sort of results were assumed known? Unique prime factorization? – Tobias Kildetoft Jul 08 '13 at 12:51
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1possible duplicate of How can I prove that $\gcd(a,b)=1\implies \gcd(a^2,b^2)=1$ without using prime decomposition? --- see also http://math.stackexchange.com/questions/166839/if-gcda-b-1-then-gcdan-bn-1 – Gerry Myerson Jul 08 '13 at 12:51
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The "trick" way is to solve $ax+by =1$ then cube both sides to get a solution to $a^2X+b^2Y=1$.
The reason I call this a "trick" is that the result generalizes to any UFD, but this proof does not.
Thomas Andrews
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Because when you cube, all the terms are either divisible by $a^2$ or $b^2$. If you square, you are left with one term $2abxy$ which is not divisible by either, so you don't get a solution for $a^2X+b^2Y=1$. – Thomas Andrews Jul 08 '13 at 13:05
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Squaring, though, can be used to show that $(a,b)=1\implies (a^2,b)=1$. And then $(b,a^2)=1\implies (b^2,a^2)=1$. So you could go that route. Just requires an extra step. – Thomas Andrews Jul 08 '13 at 13:07
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For example: $1=(ax+by)^2 = a^2x^2 + 2axby + b^2y^2 = a^2(x^2) + b(2axy+by^2)$. Setting $X=x^2$ and $Y=2axy+by^2$, then $1=a^2X+bY$. So $(a^2,b)=1$. Again, this is the "trick" answer, it is better to try to understand the other answers. – Thomas Andrews Jul 08 '13 at 13:12
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Let's think about the prime factors of $a$ and $b$. $\gcd(a,b)=1$ means that there's no common prime factor of $a$ and $b$. Now when we square $a$ and $b$ they have the same prime factors as before but twice all of them. As $a$ and $b$ has no common prime factor neither does $a^2$ and $b^2$. So $gcd(a^2,b^2)=1$ hence PROVED :)
