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How to prove $\gcd(a^2, b^2) = (\gcd(a, b))^2$?

My attempt:

Let $\gcd(a, b) = d$.

Then $d|a$ and $d|b$ then $d^2|a^2$ and $d^2|b^2$. i.e $d^2$ divides $a^2 ~~\&~~ b^2$.

Robert Soupe
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user1942348
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5 Answers5

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Hint: Factor $a$ and $b$ into primes. If $a=p_1^{n_1}\\cdot\ldots\cdot p_k^{n_k}$ and $b=p_1^{m_1}\cdot\ldots\cdot p_k^{m_k}$ with each $n_i,m_i\ge 0$. Then

$$(a,b)=p_1^{\min\{n_1,m_1\}}\cdot\ldots\cdot p_k^{\min\{n_k,m_k\}}.$$

Tim Raczkowski
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  • Please explain in details. – user1942348 Mar 21 '15 at 18:40
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    If you square a number this amounts to doubling all the exponents in its prime factorization. $\min(2x,2y) = 2\min(x,y)$ so this proves it. You can prove virtually any fact about $\gcd$ using this method; it's usually much easier and more direct. – Ibrahim Tencer Mar 21 '15 at 21:06
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$\begin{align}(a,b)(a^2,b^2) =&\ (a^3,a^2b,ab^2,b^3)\quad [\text{by basic gcd laws - see Remark below}]\\ =&\ (a,b)^3\end{align}$

Thus $\ (a^2,b^2) = (a,b)^2\ $ by cancelling $\,(a,b) \neq 0.\ \ $ [See here for the cubic analog]

Or: Gauss's Lemma (GL) yields a slick proof. Let $\rm\:{\cal C}(f)\:$ denote the content of a polynomial, i.e. the gcd of its coefficients. Then GL $\rm\: \Rightarrow\ {\cal C}(f\:g) \color{#c00}{\overset{\rm GL_{\phantom |}\!}=}\ {\cal C}(f)\ {\cal C}(g)\ $ so

$$ (a,b)^2\! =\ {\cal C}\:(ax\! +\! b)\ {\cal C}\:(a x\! -\! b)\, \color{#c00}{\overset{\rm GL_{\phantom |}\!}=}\, {\cal C}\:((a x\! +\! b)(ax\! -\! b))\, =\, {\cal C}\:(a^2 x^2\! -\! b^2)\, =\, (a^2,b^2)\qquad$$

Replying to comments, we used only basic gcd laws (associative & commutative & distributive), e.g. let's do the gcd analog of the binomial theorem (in detail)

$\ (a,b)^2\! = (a,b)(a,b) = ((a,b)a,(a,b)b) = ((a^2,ba),(ab,b^2)) = (a^2,ba,ab,b^2) = (a^2,ab,b^2)$

Similarly $\,(a,b)^3 = (a,b)(a,b)^2 = (a,b)(a^2,ab,b^2) =\,\cdots\, = (a^3,a^2b,ab^2,b^3)$

By induction $\ (a,b)^n = (a^n, a^{n-1}b,\cdots\,ab^{n-1},b^n)\ $ is straightforwardly proved.

Such gcd arithmetic requires no ingenuity, it is analogous to multiplying polynomials.

The proof generalizes to the Freshman' Dream $\,(a,b)^n = (a^n,b^n)\,$ for gcds and invertible ideals.

Bill Dubuque
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    How is $(a^3,a^2b,ab^2,b^3)=(a,b)^3$ any more obvious than $(a^2,b^2)=(a,b)^2$? – Barry Cipra Mar 21 '15 at 19:05
  • @Barry It's just gcd binomial expansion using basic gcd laws (associative, commutative, distributive). The latter is not "obvious" since $,(a,b)^2 = (a^2,ab,b^2).,$ It requires proof to show it equals $,(a^2,b^2).\ \ $ – Bill Dubuque Mar 21 '15 at 19:11
  • Ah, thanks! I was just being stoneheaded. – Barry Cipra Mar 21 '15 at 19:16
  • @Barry I elaborated a bit in a remark. – Bill Dubuque Mar 21 '15 at 19:33
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    But isn't it a theorem (or lemma, if you like) in its own right that $(a,b)(c,d)=(ac,ad,bc,bd)$? It's easy enough to show, based purely on the definition of greatest common divisor, that $(a,b)(c,d)$ divides $(ac,ad,bc,bd)$. But showing the other direction, it seems to me, requires additional work. – Barry Cipra Mar 21 '15 at 20:09
  • On further thought, it occurs to me even the "easy" direction in my previous comment requires a little lemma about gcds, namely that if $d\mid a$ and $d\mid b$, then $d\mid(a,b)$. – Barry Cipra Mar 21 '15 at 20:12
  • @Barry It's not a Theorem but, rather, just a consequence of trivial gcd arithmetic. As I show in the Remark, it follows by rote gcd arithmetic using said gcd laws (most notably the gcd Distributive Law). The calculation of $,(a,b)^3,$ is precisely analogous (and just as trivial) as the calculation of $,(a+b)^3,$ using the analogous ring laws / axioms (associative, commutative, distributive). – Bill Dubuque Mar 21 '15 at 20:13
  • @Barry In fact, if you write $,\gcd(a,b) = a\dot+b,$ then the arithmetic looks just like ring arithmetic, except we have the additional law that $, a\dot+ab = a.,$ See here and here for some examples. – Bill Dubuque Mar 21 '15 at 20:29
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    Bill, many thanks for the additional replies. I agree, it's all ultimately trivial gcd arithmetic, but that's sort of my point: We don't know which, if any, of these easily proved points the OP has at his or her disposal. – Barry Cipra Mar 21 '15 at 20:40
  • @Barry The (easy) proofs of the basic gcd laws are all on MSE (and in most textbooks). But, alas, rarely do elementary textbooks emphasize the analogy with integer (ring) arithmetic. No doubt that it is difficult to judge which proofs will be most accessible to OPs, since they usually do not provide enough context to infer that. – Bill Dubuque Mar 21 '15 at 21:00
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If $\gcd(a,b)=d$, then
$a = a'd$
$b = b'd$
and $\gcd(a',b') = 1$

That means
$a^2 = {a'}^2d^2$
$b^2 = {b'}^2d^2$
and $\gcd({a'}^2,{b'}^2) = 1$ $ ~~\color{red}{ \star }$

Hence $\gcd(a^2,b^2)=d^2$.

$ ~~\color{red}{ \star }$ pelase see this.

Community
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AgentS
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I'll use the notation of Michael's answer.

Since $g_2\mid a^2$, $g_2\mid b^2$, $g_1\mid a$ and $g_1\mid b$, there are integers $p,q,r,s$ such that:

$$a^2 = g_2p, \ b^2=g_2q, \ a=g_1r, \ b=g_1s \ ,$$

with $gcd(p,q)=1$ and $gcd(r,s)=1$. Manipulating those expressions, is easy to get that $r^2q=s^2p$ and, by Euclid's lemma, $r^2\mid p$ and $p\mid r^2$ thus $p=r^2$ (assuming positive integers), similarly, $q=s^2$ so finally

$$g_1^2 = \frac{a^2}{r^2}=\frac{g_2p}{r^2} = g_2$$

Here I use that if $gcd(r,s)=1$, then $gcd(r^2,s^2)=1$. This is much easier to prove than the general case so my reasoning is not circular.

Daniel
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Hint (for the easy direction): Suppose that $g_1=\gcd(a,b)$ and $g_2=\gcd(a^2,b^2)$. Since $g_1\mid a$ and $g_1\mid b$, then $g_1^2\mid a^2$ and $g_1^2\mid b^2$. Therefore, by the definition of the greatest common divisor, $g_1^2\mid g_2$.

Michael Burr
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