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

Question

Let $a$ and $b$ be two integers. Show that $\gcd(a^2, b^2) = \gcd(a,b)^2$.

This is what I have done so far:

Let $d = \gcd(a,b)$. Then $d=ax+by$ for some $x,y$. Then $d^2 =(ax+by)^2 = a^2x^2 + 2axby+b^2y^2$.

I am trying to create a linear combination of $a^2$ and $b^2$ but do not know what to do with the middle term.

EDIT: I would be interested in seeing a proof using the Fundamental Theorem of Arithmetic as well. I simply thought it would be easiest to use Bezout's Identity.

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

Answer 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 $2$.

Answer 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 $(a^3,a^2b,ab^2,b^3)$. $\ \ \ \square$

Answer 4

We extend the case to which 2 is replaced by any natural number, that is, we're proving that $$\forall n \in \mathbb{N}, \; \text{gcd(a$^n$,b$^n$)}=\text{gcd(a,b)$^n$}$$

By Benzout's lemma, \begin{align*}\exists p,q \in \mathbb{Z}: \: \gcd(a^n,b^n)&=a^np+b^nq \\ &= \gcd(a,b)^n \left(\left(\frac{a}{\gcd(a,b)}\right)^np + \left(\frac{b}{\gcd(a,b)}\right)^nq \right) \end{align*} $$\implies \gcd(a,b)^n \bigr| \gcd(a^n,b^n)$$

Let $ \: d=\text{gcd(a,b)}$
Again, by benzout's lemma, $$\exists x,y \in \mathbb{Z}: \, d=ax+by $$ \begin{align*} d^{2n-1} &= \sum_{k=0}^{2n-1} \binom{2n-1}{k} x^{2n-1-k}y^k \\ &= \sum_{k=0}^{n-1} \binom{2n-1}{k} x^{2n-1-k}y^k + \sum_{k=n}^{2n-1} \binom{2n-1}{k} x^{2n-1-k}y^k \end{align*} Substitute $k \mapsto 2n-1-k $ at the second sum \begin{align*} d^{2n-1} &=\sum_{k=0}^{n-1} \binom{2n-1}{k} x^{2n-1-k}y^k \, + \, \sum_{k=0}^{n-1} \binom{2n-1}{k} x^{k}y^{2n-1-k} \\ &=x^n \sum_{k=0}^{n-1} \binom{2n-1}{k} x^{n-1-k}y^k \, + \, y^n \sum_{k=0}^{n-1} \binom{2n-1}{k} x^k y^{n-1-k} \\ d^n &= x^n \left( \sum_{k=0}^{n-1} \binom{2n-1}{k} \left(\frac{x}{d}\right)^{n-1-k} \left(\frac{y}{d} \right)^k \right) + y^n \left(\sum_{k=0}^{n-1} \binom{2n-1}{k} \left(\frac{x}{d}\right)^k \left(\frac{y}{d}\right)^{n-1-k} \right) \end{align*} \begin{align*} &\implies \exists u,v \in \mathbb{Z} : \: d^n = a^nu +b^n v \\ &\implies \gcd(a^n,b^n) \, \bigr| \, d^n=\gcd(a,b)^n \end{align*} Both divide each other $$\therefore \; \gcd(a^n,b^n)=\gcd(a,b)^n $$