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

$\gcd(a,b)=d\Rightarrow\ d\mid a,b\Rightarrow\ \ d^2\mid a^2,b^2\Rightarrow\ d^2\mid\gcd(a^2,b^2)$.
But to complete the proof we must show that: $\gcd(a^2,b^2)\mid d^2$
How can I achieve this?

Martin Sleziak
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Hamid Reza Ebrahimi
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  • Also special case of this one: http://math.stackexchange.com/questions/524454/how-to-prove-greatest-common-divisor-using-bézouts-lemma. – Martin R Apr 14 '16 at 15:19

3 Answers3

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Write $a=da'$ and $b=db'$ with $\gcd(a',b')=1$.

Then $$ \dfrac{a^2}{b^2}=\dfrac{d^2(a')^2}{d^2(b')^2}=\dfrac{(a')^2}{(b')^2} $$ which cannot be further reduced.

lhf
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Simply, write $a=da'$ and $b=db'$ such that $$\gcd(a', b') =1$$ Hence, $a^2= d^2(a')^2$ and $b^2=d^2(b')^2$.

Since, $\gcd(a', b') =1$, $\gcd((a')^2, (b')^2) =1$ and thus $$gcd(a^2, b^2) = d^2\gcd((a')^2, (b')^2) = d^2$$

Aritra Das
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  • And how to be sure that: $\gcd(m,n)=1\Rightarrow\gcd(m^2,n^2)=1$ ? – Hamid Reza Ebrahimi Apr 14 '16 at 15:04
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    @HamidRezaEbrahimi Since $m,n$ are coprime, they have no common prime factor. Hence their squares will also not contain any common prime factors since squaring simply doubles the power of each prime factor and doesn't introduce any new prime factor. Thus their squares are also coprime. – Aritra Das Apr 14 '16 at 15:06
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By dividing $d$ on both sides and using $\gcd(ac,bc)=c\gcd(a,b)$, we may assume that $\gcd(a,b)=d=1$, so you only need to show that if $\gcd(a,b)=1$, then $\gcd(a^2,b^2)=1$. Suppose that $p$ is a prime that divides $a^2$ and $b^2$, then $p$ divides $a$ and $b$, this contradict to the assumption $\gcd(a,b)=1$, so there is no prime divides $a^2$ and $b^2$, it follows that $\gcd(a^2,b^2)=1$.

Xiang Yu
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