n3+3n2+5n+3=(n+1)(n2+2n+3) =(n+1)(n+1)2+2] Consider the following cases n=3k,n3+3n2+5n+3 =(3k+1)[(3k+1)2+2] =(3k+1)[9k2+6k+3], which is divisible by 3 n=3k+1, n3+3n2+5n+3 =(3k+2)(9k2+12k+4+2), =(3k+2)(9k2+12k+4+2), which is divisible by 3 n=3k+2 n3+3n2+5n+3=(3k+3)(3k+3)2+2 which is divisible by 3 In all the above cases n3+3n2+5n+3 is divisible by 3 ∴n3+3n25n+3 is divisible by 3 for all n∈Z ∴S=ϕ