Show that $\gcd(a^2, b^2) = \gcd(a,b)^2$

Solution 1:

Suppose that $(a,b)=d$. Then Bezout's Identity says that we have some $x,y$ so that $ax+by=d$, and therefore $$ a^3x^3+3a^2bx^2y+3ab^2xy^2+b^3y^3=d^3\tag{1} $$ Dividing $(1)$ by $d$, remembering that both $d\mid a$ and $d\mid b$, we get $$ a^2\left(\frac adx^3+3\frac bdx^2y\right)+b^2\left(3\frac adxy^2+\frac bdy^3\right)=d^2\tag{2} $$ $(2)$ says that $\left.\left(a^2,b^2\right)\middle|\,d^2\right.$. Since $d\mid a$ and $d\mid b$, we also have $\left.d^2\,\middle|\left(a^2,b^2\right)\right.$.

Therefore, $$ \left(a^2,b^2\right)=d^2=(a,b)^2 $$

Solution 2:

Hint: Greatest common divisor will always have the minimum of the exponents(from the $2$ number) from the prime factorization. The minimum exponents of $a^2$ and $b^2$ are the same like the minimal exponents of $a$ and $b$ multiplied by $2$.

Solution 3:

$(a,b)$ is the usual notation for $\gcd(a,b)$.

Remember $\, n\mid a,b\iff n\mid (a,b)\,$ by definition of $\gcd$.

Use distributive property $\,(ak,bk)=k(a,b)$.

As Bill says, you can prove it in this similar way:

$c\mid (ak,bk)\iff c\mid ak,bk\iff \frac{c}{k}\mid a,b\iff\frac{c}{k}\mid (a,b)\iff c\mid k(a,b)$

Another lemma: $((a,b),c)=(a,b,c)$.

Proof: $(a,b,c)\mid a,b,c\iff (a,b,c)\mid (a,b),c\iff (a,b,c)\mid ((a,b),c)$

$((a,b),c)\mid (a,b),c\iff ((a,b),c)\mid a,b,c\iff ((a,b),c)\mid (a,b,c)\ \ \ \square$

Prove $(a,b)(a^2,b^2)=(a,b)^3$, then divide by $(a,b)$ to finish your proof.

$(a,b)(a^2,b^2)=(a(a^2,b^2),b(a^2,b^2))=((a^3,ab^2),(a^2b,b^3))=(a^3,a^2b,ab^2,b^3)$

$(a,b)(a,b)=((a,b)a,(a,b)b)=((a^2,ab),(ab,b^2))=(a^2,ab,b^2)$

$(a^2,ab,b^2)(a,b)=(a^2(a,b),ab(a,b),b^2(a,b))=((a^3,a^2b),(a^2b,ab^2),(ab^2,b^3)),$

which too is $(a^3,a^2b,ab^2,b^3)$. $\ \ \ \square$