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

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

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?

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## marked as duplicate by Cameron Buie, Andrés E. Caicedo, Jared, Amzoti, Davide GiraudoAug 9 '13 at 16:09

$a^2$ and $b^2$ have the same prime factors as $a$ and $b$ – AlexM Aug 8 '13 at 23:22

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

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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.

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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 9 '13 at 0:17
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$. – Thomas Andrews Aug 9 '13 at 4:37 Why is this less informative? It shows explicitly that appeals to prime factorization are unnecessary, and provides us with an algorithm to verify the result (by exhibiting witnesses$t,w$such that$a^2t+b^2w=1$). – Andrés E. Caicedo Aug 9 '13 at 6:38 For a first number-theory course, the multiplicative structure of the natural numbers has a more concrete feel. Bezout's theorem may by contrast feel like magic, until one internalizes its algebraic content. – André Nicolas Aug 9 '13 at 6:43 gcd$(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$. - If$\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\$.

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