6a+1 \(\vdots\) 3a+1
=> \(\dfrac{6a+1}{3a+1}=\dfrac{3a+3+1}{3a+1}=\dfrac{3a+1}{3a+1}+\dfrac{3}{3a+1}=1+\dfrac{3}{3a+1} \) \(\in Z\)
mà \(1 \in Z\)
=> \(\dfrac{3}{3a+1} \in Z\) => \(3a+1 \in \) Ư(3) = { -3 ; -1 ; 1 ; 3 }
Ta có bảng sau :
3a+1 | -3 | -1 | 1 | 3 |
3a | -4 | -2 | 0 | 2 |
a | \(-\frac{4}{3}\) | \(-\frac{2}{3}\) | 0 | \(\frac{2}{3}\) |
Vì a là số nguyên => a=0
Vậy..