[sompong2479]

Easy Problem:
Let d=gcd(a,b). Then
\[
\frac{a}{d}\Big|\left(\frac{b}{d}\right)^2d\Longrightarrow\frac{a}{d}\Big|d.
\]
Similarly, get \((a/d)^3|d,(a/d)^5|d,\ldots\) and so on. Hence \( a=d\), that is \( a|b \). Let \( b=ka\). Then one has from the \( 2,4,6,\ldots \)'th equations that
\[
k^2|a,k^4|a,\ldots
\]
and so on. This is the case if and only if \( k=1\). Done.