Squares of two coprime numbers

Question

If there are two coprime numbers a and b, then are a^2 and b^2 also coprime ?

Yes, they are. Any common prime factor of $a^2$ and $b^2$ would also divide $a$ and $b$. — , Jan 14 '18 at 16:10
Alternatively, here is the proof of a more general statement: If $\gcd(a,b)=1$, then $\gcd(a^n,b^n)=1$ — projectilemotion, Jan 14 '18 at 16:37

score 2 · Answer 1 · answered Jan 14 '18 at 16:18

Yes. You can see that two ways:

$a$ and $b$ coprime means they share no prime factor. As the prime factors of $n^2$ are exactly the prime factors of $n$, but squared, we see $a^2$ and $b^2$ share no prime factors either.
Via Bézout's identity, we show that, if $a$ and $b$ are coprime, $a^2$ and $b$ are too (and hence $a^2$ and $b^2$ are).

Indeed, coprimality of $a$ and $b$ means there exist integers $u$ and $v$ such that $ua+vb=1$. Squaring this relation, we obtain $$1=u^2a^2+2uvab+v^2b^2=u^2a^2+(2uva+v^2b)b.$$

Yes, it's a fine way – maybe less natural, and harder to generalise to proving $a^i$ and $b^j$ are coprime for any $i,j$. — Bernard, Jan 14 '18 at 16:38

score 1 · Accepted Answer · answered Jan 14 '18 at 16:21

Yes the squares are also coprime.

Note that the prime factors of $a^2$ are the same as prime factors of $a$but the power of each prime in $a^2$ is twice the power of the same prime in $a$.

If $a$ and $b$ do not share a prime then $a^2$ and $b^2$ do not share any prime as well

Squares of two coprime numbers

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