Suppose: n² + n + 2 ⋮15

=> n(n+1) + 2 ⋮ 15 

But n(n+1) ⋮ 2 => n(n+1) + 2 is an even number

=> n(n+1) + 2 has last digit is 0

=> n(n+1) has last digit is 8

But the last digit of product of two consecutive integers is 0,2,6 

So n(n+1) + 2 \(⋮̸15\)

So n(n+1) + 2 \(⋮̸3\)

=> n(n+1) + 2 - 3 = n² + n - 1 \(⋮̸3\)

 => \(\left(n^2+n-1\right)^2\equiv1\left(mod3\right)\Rightarrow\left(n^2+n-1\right)^2-1⋮3\)

\(\left(n^2+n-1\right)^2-1=\left(n^2+n-2\right)\left(n^2+n\right)\)is the profuct of two consecutive even number so has the form 2k.(2k+2) = 4k² + 4k = 4k(k+1) \(⋮8\)

\(\left(3,8\right)=1\Rightarrow\left(n^2+n-1\right)^2-1⋮24\) 

We have: \(\left(n^2+n-1\right)^2-1=\left(n^2+n\right)\left(n^2+n-2\right)=\left(n-1\right)n\left(n+1\right)\left(n+2\right)⋮24\left(n\in Z\right)\)                                                                                                                         (product of four consecutive numbers)

So ...