Four integers from three pairwise products

Integers $a,b,c,d$, each greater than $1$, satisfy $ab=21$, $bc=15$ and $cd=45$. What is $a^2+b^2+c^2+d^2$?

164. $b$ divides both $21$ and $15$ and exceeds $1$, so $b=3$. Then $a=7$, $c=5$ and $d=9$, and $49+9+25+81=164$.