Ta có :
`(3n+1)^2 - 25`
`= (3n+1)^2 - 5^2`
`= (3n+1+5)(3n+1-5)`
`= (3n+6)(3n-4)`
` = 3 (n+2)(3n-4)`
`\forall n \in ZZ` thì `3(n+2)(3n-4) \in ZZ`
Mà `3 \vdots 3` nên `3 (n+2)(3n-4) \vdots 3`
`=> (3n+1)^2- 25 \vdots 3`
Vậy `(3n+1)^2-25 \vdots 3 \forall x \in NN`