$\gcd(m^2,n^2)$ = $(\gcd(m,n))^2$

Question

Let $\gcd(m,n)=d$ and $\gcd(m^2,n^2)=d'$.

It means $d|m$ and $d|n$ which implies $d^2|m^2$ and $d^2|n^2$ Thus $d^2|d'$. How to prove reverse $d'|d^2$

Answer 1

Another way:

Let $\dfrac nN=\dfrac mM=d$

so that $(M,N)=1$

In that case, $N^2,M^2$ cannot share a common factor $>1$

$\implies(M^2,N^2)=1$

Method$\#:2$

Let the highest exponent of prime $p$ that divides $m,n$ be $a,b$ respectively

So, the highest exponent of $p$ in $(m,n)$ will be min$(a,b)=c$(say)

Similarly the highest exponent of prime $p$ in $(m^2,n^2)$ will be min$(2a,2b)=2c$

This will hold true for any prime that divides $mn$

Answer 2

Perhaps overkill but a handy concept is that for any $m,n$ there are $d =\gcd(m,n)$ and $n', m'$ so that $m = dm'$ and $n = dn'$ and it's not difficult to prove $m', n'$ are relatively prime.

(Because if $\gcd(n', m') = k>1$ then $kd|n$ and $kd|m$ and $kd > d$ so $d\ne \gcd(m,n)$ after all.)

SO let $m = dm'$ and $n = dn'$. Then $m^2 = d^2m'^2$ and $n^2 = d^2n'^2$. $d^2|m^2, n^2$ but then $m'^2$ and $n'^2$ have no factors in common. So there is no common divisor greater than $d^2$ that $d^2$ divides. So $d^2$ is greatest common divisor.

(Remember the Theorem: if $k|m,n$ then $k|\gcd(m,n)$.