If there are two coprime numbers a and b, then are a^2 and b^2 also coprime ?
2026-02-22 17:54:05.1771782845
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Squares of two coprime numbers
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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 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