`\qquad f(x)=ax^2+bx+c` `(a;b;c\in ZZ; a\ne 0)`
Vì `f(3)` chia hết cho $9$
`=>(a.3^2+b.3+c)\ \vdots\ 9`
`=>(9a+3b+c)\ \vdots\ 9`
Vì `9a\ \vdots\ 9`
`=>(3b+c)\ \vdots\ 9`
`=>8.(3b+c)\ \vdots\ (8.9)`
`=>(24b+8c)\ \vdots\ 72` $(1)$
$\\$
Vì `f(4)` chia hết cho $8$
`=>(a.4^2+b.4+c)\ \vdots\ 8`
`=>(16a+4b+c)\ \vdots\ 8`
Vì `16a\ \vdots\ 8`
`=>(4b+c)\ \vdots\ 8` (*)
`=>9.(4b+c)\ \vdots\ (9.8)`
`=>(36b+9c)\ \vdots\ 72` $(2)$
$\\$
Từ `(1);(2)=>(36b+9c-24b-8c)\ \vdots\ 72`
`=>(12b+c)\ \vdots\ 72`
$\\$
Ta có:
`f(12)=a.12^2+b.12+c`
`=144a+12b+c`
Vì `144a\ \vdots\ 72` và `(12b+c)\ \vdots\ 72`
`=>f(12)=144a+12b+c \ \vdots\ 72`
Vậy `f(12)` chia hết cho ` 72`
