957

若 3 个整数 ${\displaystyle a,b,c}$ 满足 ${\displaystyle a^2+b^2+c^2+3<ab+3b+3c}$, 则这样的有序整数组 ${\displaystyle (a,b,c)}$ 共有_____组.

配方得$$\begin{array}{r}(2a-b{)}^2+3(b-2{)}^2+(2c-3{)}^2<9.\end{array}$$
那么至少 ${\displaystyle 3(b-2{)}^2<9\Rightarrow b=1,2,3}$.

• ${\displaystyle b=1}$, 则 ${\displaystyle (2a-1{)}^2+3+(2c-3{)}^2<9}$, ${\displaystyle (a,c)=(0,1),(0,2),(1,1),(1,2)}$.
• ${\displaystyle b=2}$, 则 ${\displaystyle (2a-2{)}^2+(2c-3{)}^2<9}$, ${\displaystyle (a,c)=(0,1),(0,2),(1,1),(1,2),(2,1),(2,2)}$.
• ${\displaystyle b=3}$, 与 ${\displaystyle b=1}$ 的解数相同.

这样, 有序的 ${\displaystyle (a,b,c)}$ 有 ${\displaystyle 14}$ 个.