`ab+a+b+2=(a+1)(b+1)+1 \vdots 24`

`@` `(a+1)(b+1)+1 \vdots 3`

`=> (a+1)(b+1) \equiv 2 [3]`

`=>a+1` hoặc `b+1` đồng dư `1` khi mod `3`

`=> a` hoặc `b` chia hết cho `3`

`=> ab` chia hết `3`

`@` `(a+1)(b+1)+1 \vdots 8`

`=> (a+1)(b+1) \equiv 7 [8]`

$\left[\begin{matrix} \left[\begin{matrix} \begin{cases} a+1 \equiv 7 [8]\\b+1 \equiv 1 [8]\end{cases}\\ \begin{cases} a+1 \equiv 1 [8]\\b+1 \equiv 7 [8] \end{cases}\end{matrix}\right.\\\left[\begin{matrix} \begin{cases} a+1 \equiv 3 [8]\\b+1 \equiv 5 [8]\end{cases}\\ \begin{cases} a+1 \equiv 5 [8]\\b+1 \equiv 3 [8] \end{cases}\end{matrix}\right.\end{matrix}\right.$

$\left[\begin{matrix} \left[\begin{matrix} b \equiv 0[8]\\a \equiv 0[8] \end{matrix}\right.\\\left[\begin{matrix} \begin{cases} a \equiv 2 [8]\\b \equiv 4 [8]\end{cases}\\ \begin{cases} a \equiv 4 [8]\\b \equiv 2 [8] \end{cases}\end{matrix}\right.\end{matrix}\right.$

$\left[\begin{matrix} \left[\begin{matrix} b \equiv 0[8]\\a \equiv 0[8] \end{matrix}\right.\\\left[\begin{matrix} \begin{cases} a \vdots 2 \\b \vdots 4 \end{cases}\\ \begin{cases} a \vdots 4 \\b \vdots 2 \end{cases}\end{matrix}\right.\end{matrix}\right.$

`=> ab \vdots 8`

Mà `(3;8)=1` nên `ab \vdots 24`

`=> đpcm`