The given $2^{nd}$ relation can be simplified to $$b=4ac^2-4c^4=4c^2(a-c^2)$$ Therefore, we find $$4c^2 | b\implies 450 > b \ge 4c^2 \implies c \le 11 $$ Meanwhile, setting $b=4ac^2 -4c^4$ to $a+b+c=450$ yields $$a=c^2 + \frac{450-c-c^2}{1+4c^2}$$ Trying $c=1, 2, \cdots, 10$ finds one qualified solution $$(a,b,c)=\boxed{(7, 51, 392)}$$