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So, if $\gcd(a,b)=1$, then $\gcd(a^2,b^2)=1$ means $1=ax+by$, and want to show $a^2x+b^2y=1$.

By squaring $1=ax+by$ both sides, I get, $1=(ax)^2+2(ax)(by)+(by)^2$, but this doesn't help my proof.

JMP
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Wes
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    Try cubing it. ${}$ – Daniel Fischer Oct 03 '13 at 02:32
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    Ok, I cubed it. And got, $a^2(ax+3bx^2y)+b^2(3axy^2+by^3)$. So I got what I want I think. But Why cube helps? I don't understand that part.. Why to cube works? – Wes Oct 03 '13 at 02:38
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    Because it produces the right powers. Now generalize to $a^n, b^n$. – André Nicolas Oct 03 '13 at 02:41
  • If one uses the algebro-number-theoretic definition of "prime" (which definition is "morally" or pedagogically correct is debatable I suppose), there is a nice proof as pointed out by T.Bongers: if $(a^2,b^2)>1$ then $p\mid a^2,b^2\Rightarrow p\mid a,b\Rightarrow p\mid (a,b)=1$ for some prime $p$, absurd. – anon Oct 03 '13 at 02:45
  • @ Andre What do you mean 'generalize'? Like, making a formula? – Wes Oct 03 '13 at 02:46
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    please convince yourself that $x,y$ in $ax+by=1$ and $x,y$ in $a^2x+b^2y=1$ are not the same... I mean, dont try to prove $ax+by=1$ implies $a^2x+b^2y=1$... This should help you i guess... –  Oct 03 '13 at 02:46
  • I'm not allowed to use prime yet. – Wes Oct 03 '13 at 02:48
  • See also http://math.stackexchange.com/questions/63323/how-to-use-fundamental-theorem-of-arithmetic-to-conclude-that-gcdak-bn-1. – lhf Oct 03 '13 at 03:03
  • @lhf : please consider above comment, he is not allowed to use prime yet. So there is no question of considering fundamental theorem of arithmetic.. So, i consider this is not a duplicate.. –  Oct 03 '13 at 03:06

6 Answers6

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Suppose $gcd(a,b)=1$ then you have $ax+by=1$, cubing this, we get $(ax+by)^3=1$

i.e., $a^3x^3+b^3y^3+3a^2x^2by+3axb^2y^2=1$

i.e., $a^2(ax^3+3x^2by)+b^2(by^3+3axy^2)=1$

does this imply $gcd(a^2,b^2)=1$???

9

First show $\gcd(a, b^2) = \gcd(a^2, b) = 1$

$\gcd(a,b) = 1$ means that there are $x$, $y$ ($x$, $y$ are integers) such that $ax + by =1$

$ax + by(ax+by) =1$

$a(x+bxy) + b^2y^2 =1$

That means $\gcd(a,b^2)=1$

By a similar derivation, $\gcd(a^2,b)=1$.

$ar_1(ar_2+b^2s_2) +b^2s_1 = 1$

$a^2r_1^2+b^2(ar_1s_1+s_1)=1$

Therefore,

$\gcd(a^2, b^2) = 1$

Ted Hopp
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chu
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Cube it! ${}{}{}{}{}{}{}{}{}{}{}{}{}{}$

André Nicolas
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You might also try a contradiction:

Suppose $(a,b) = 1$ and $(a^2,b^2) = d$, where $d \neq 1$. Using the second assumption, you can say $a^2x + b^2y = d$ for some pair of integers $x$ and $y$. Dividing both sides by $d$ can lead to a contradiction with $(a,b) = 1$.

There are of course some details missing, but this is worth considering, as cubing may not be very natural.

Dannum
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suppose $gcd(a^2, b^2) = d$ and $d \ne 1$

Then $a^2 = dm, b^2 = dn \implies a^2b^2 = d^2mn$, where $gcd(m,n) = 1$

This implies both $m$ and $n$ are perfect squares. $m = p^2, n = q^2$

We have $a^2 = dp^2, b^2 = dq^2 \implies d$ is a perfect square. Let $d = x^2$

Now we have $a^2 = x^2p^2, b^2 = x^2q^2 \implies a = xp, b = xq \implies gcd(a,b) = x = 1$

Hence $d = x^2 = 1$

sku
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As $1 = ax + by$, then $ab = a^2bx + ab^2y$.

Therefore,

$1 = (ax + by)^2 = a^2x + b^2y + 2abxy$

$= a^2x + b^2y + 2(a^2bx + ab^2y)xy$

$= a^2(x + 2bx^2y) + b^2(y + 2axy^2)$

JMP
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