How do I show that $\gcd(a^2, b^2) = 1$ when $\gcd(a,b)=1$?

Question

I can show that $\gcd(a,b)=1$ implies $\gcd(a^2,b)=1$ and $\gcd(a,b^2)=1$. But what do I do here?

$a^2$ and $b^2$ have the same prime factors as $a$ and $b$ – AlexM Aug 08 '13 at 23:22 — AlexM, Aug 08 '13 at 23:22

score 12 · Answer 1 · answered Aug 09 '13 at 00:12

12

The following proof is less informative than the posted proofs, but it is kinda cute. Recall the theorem of Bézout which says that $c$ and $d$ are relatively prime if and only if there exist integers $s$ and $t$ such that $cs+dt=1$.

Let $x$ and $y$ be integers such that $ax+by=1$. Now cube both sides. We get $$a^2(ax^3+3x^2by)+b^2(3axy^2+by^3)=1,$$ and therefore by the theorem of Bézout $a^2$ and $b^2$ are relatively prime.

answered Aug 09 '13 at 00:12

André Nicolas

514,336

2

That is cute! Very cool. (We can't just square, of course, because the $abxy$ cross term has first powers of both $a$ and $b$ in it, but going to the cube ensures that every term has at least one of $a$ and $b$ raised to the second power.) This would generalize to all higher powers by just taking the right power of the equation $ax+by=1$ too... – Steven Stadnicki Aug 09 '13 at 00:17
Yes, there are whole slews of GCD theorems for which Bezout gives a quick proof without arguments about prime factorizations. You can square to prove that $(a,b)=1\implies (a,b^2)=1$. And then note the same proof shows taht $(b^2,a)=1\implies $(b^2,a^2)=1$. – Thomas Andrews Aug 09 '13 at 04:37
Why is this less informative? It shows explicitly that appeals to prime factorization are unnecessary, and provides us with an algorithm to verify the result (by exhibiting witnesses $t,w$ such that $a^2t+b^2w=1$). – Andrés E. Caicedo Aug 09 '13 at 06:38
For a first number-theory course, the multiplicative structure of the natural numbers has a more concrete feel. Bezout's theorem may by contrast feel like magic, until one internalizes its algebraic content. – André Nicolas Aug 09 '13 at 06:43

score 10 · Accepted Answer · answered Aug 08 '13 at 23:22

10

Hint: You've shown that $\gcd(y,b^2)=1$ when $\gcd(y,b)=1.$ What happens when $y=a^2$?

answered Aug 08 '13 at 23:22

Cameron Buie

105,149

score 5 · Answer 3 · answered Aug 08 '13 at 23:28

5

gcd$(a,b)=1$ if and only if no prime divides a and b. A prime divides $a^2$ if and only if it divides a. Therefore a number divides $a^2$ and $b^2$ if and only if it divides $a$ and $b$.

answered Aug 08 '13 at 23:28

Asinomás

107,565

score 0 · Answer 4 · answered Aug 09 '13 at 06:32

If $\gcd(a, b) = 1$, this means exactly that $a$ and $b$ do not have any prime factors in common. Any common factor would constitute a common divisor. If $a$ and $b$ have some prime factors in common, then the product of those common factors (including any repeated ones of higher multiplicity) is the greatest common divisor.

For instance $24 = 2 \cdot 3\cdot 4$ and $60 = 3\cdot 4\cdot 5$. The common factors are $3$ and $4$ and so $\gcd(24, 60) = 3\cdot 4 = 12$.

If we take an integer $a$ and square it, the resulting integer $a^2$ does not have any new prime factors which are not already present in $a$. It simply has the same factors, with double the multiplicity.

For instance $12 = 3\cdot 2\cdot 2$, and $144 = 12^2 = (3\cdot 2\cdot 2)(3\cdot 2 \cdot 2) = 3\cdot 3\cdot 2\cdot 2\cdot 2\cdot 2$.

So if $a$ and $b$ have no prime factors in common, then $a^2$ and $b^2$ have no prime factors in common either, and so $\gcd(a^2, b^2) = 1$.

How do I show that $\gcd(a^2, b^2) = 1$ when $\gcd(a,b)=1$?

4 Answers4