Prove that if $\gcd(a,b)=1$, then $\gcd(a^2,b^2)=1$ 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.
 A: 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,
$\gcd(a^2, b^2) = 1$
A: Cube it!   ${}{}{}{}{}{}{}{}{}{}{}{}{}{}$
A: 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 $gcd(a^2,b^2)=1$???
A: 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.
A: 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 $d = x^2 = 1$
A: As $1 = ax + by$, then $ab = a^2bx + ab^2y$.
Therefore,
$1 = (ax + by)^2 = a^2x + b^2y + 2abxy$
$= a^2x + b^2y + 2(a^2bx + ab^2y)xy$
$= a^2(x + 2bx^2y) + b^2(y + 2axy^2)$
