\(\dfrac{2n+8}{n+2}=\dfrac{2\left(n+2\right)+4}{n+2}=2+\dfrac{4}{n+2}\)

\(\dfrac{2n+8}{n+2}\in N\) only when \(\dfrac{4}{n+2}\in N\) or \(4⋮n+2\)

Since \(n\in N,n\ge0\)and \(n+2\ge2\).So n + 2 = 2 or n + 2 = 4

=> n = 0 or n = 2

n+8n+2=2(n+2)+4n+2=2+4n+2

2n+8n+2∈N

 only when 4n+2∈N or 4⋮n+2

Since n∈N,n≥0

and n+2≥2

.So n + 2 = 2 or n + 2 = 4

=> n = 0 or n = 2