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

Question

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 $a^2 ~~\&~~ b^2$.

you have proved $d^2 | a^2$ and $d^2 | b^2$ but how do you know $d^2$ is the gcd($a^2, b^2)$? — sku, Sep 16 '24 at 04:32
This is exercise 1.3.18 of Niven, Zuckerman & Montgomery's An Introduction to the Theory of Numbers. — 19021605, Sep 08 '25 at 13:32

score 9 · Answer 1 · edited Sep 16 '24 at 04:17

9

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

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

edited Sep 16 '24 at 04:17

Elias Zamaria

195

answered Mar 21 '15 at 18:36

Tim Raczkowski

20,680

1

Please explain in details. – user1942348 Mar 21 '15 at 18:40
1

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

score 4 · Answer 2 · edited Apr 13 '17 at 12:21

4

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.

edited Apr 13 '17 at 12:21

Community

1

answered Mar 21 '15 at 18:42

AgentS

12,441

2

Not a good solution in my opinion because you've used the fact that if $\gcd (a',b')=1$ then $\gcd(a'^2,b'^2)=1$ this is what the question asks to prove! – Fermat Mar 21 '15 at 18:58
As $$\gcd(a,b)=d\iff \gcd({a\over d},{b\over d})=1$$ – Fermat Mar 21 '15 at 19:02
Yeah we can add that line – AgentS Mar 21 '15 at 19:04
1

It seems to me there's a tacit lemma lurking in the background here: If $(r,s)=(r,t)=1$, then $(r,st)=1$. – Barry Cipra Mar 21 '15 at 19:08
I see... let me add a reference to that result :) – AgentS Mar 21 '15 at 19:12
2

Knowing the fact that I mentioned in the first comment, the proof can be done in a line: $$\gcd(a,b)=d\implies \gcd({a\over d},{b\over d})=1\implies \gcd\big(({a\over d})^2,({b\over d})^2\big)=1\implies \gcd(a^2,b^2)=d^2$$ – Fermat Mar 21 '15 at 19:14
That looks neat indeed! – AgentS Mar 21 '15 at 19:16

Bill Dubuque · Answer 3 · 2025-11-05T08:10:44.447

4

$\begin{align}\color{#0af}{(a,b)}(a^2,b^2) =&\ (a^3,a^2b,ab^2,b^3)\quad [\text{by gcd distributive law etc - see Remark below}]\\ =&\ \color{#0af}{(a,b)}^3\end{align}$

Thus $\ (a^2,b^2) = (a,b)^2\ $ by cancelling $\,\color{#0af}{(a,b)} \neq 0.\ \ $ [here is a cubic & higher degree 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

$$\begin{align} (a,b)^2\! &=\ {\cal C}(ax\! +\! b)\ {\cal C}(a x\! -\! b)\, \\[.2em] &\,\smash[]{\color{#c00}{\overset{\rm GL_{\vphantom |}}=}}\, {\cal C}((a x\! +\! b)(ax\! -\! b))\,\\[.2em] &=\, {\cal C}(a^2 x^2\! -\! b^2)\, \\[.2em] &=\, (a^2,b^2)\end{align}$$

Remark 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's Dream $\,(a,b)^n = (a^n,b^n)\,$ for gcds and invertible ideals.

edited Nov 05 '25 at 08:10

answered Mar 21 '15 at 18:54

Bill Dubuque

285,713

1

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
1

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
1

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

score 3 · Answer 4 · answered Mar 21 '15 at 18:58

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.

score 0 · Answer 5 · answered Mar 21 '15 at 18:36

0

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, $g_1^2\mid g_2$.

answered Mar 21 '15 at 18:36

Michael Burr

34,150

1

Not useful. OP already knows this. – Anne Bauval Aug 13 '25 at 12:27

score 0 · Answer 6 · answered Nov 05 '25 at 03:16

0

The other direction can be shown using the ceiling function:

$p^k \mid a^2, b^2 \implies p^{\lceil{\frac{k}{2}\rceil}}\mid a,b \implies p^{\lceil{\frac{k}{2}\rceil}} \mid (a,b) \implies (p^{\lceil{\frac{k}{2}\rceil}})^2 \mid(a,b)^2 \implies p^k \mid (a,b)^2$

answered Nov 05 '25 at 03:16

Derek Luna

2,876

This comes from the case where $k$ is odd so that $\frac{k-1}{2} p$'s are not enough for $p^k \mid a^2$ so $a$ must have at least $\frac{k-1}{2} + 1 = \frac{k+1}{2}$ $p$'s. – Derek Luna Nov 05 '25 at 03:48

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

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