Problem: Show that the expression a(a^{2}+2)/3 is an integer for all a≥0.

**Solution:** According to division algorithm, every a is of the form 3q, 3q+1, 3q+2.

Assume the first of this cases then a(a^{2}+2)/3 = 3q{(3q)^{2}+2}/3= 3q(9q^{2}+2)/3= q(9q^{2}+2).

Which clearly is an integer.

Similarly if a = 3q+1 then

a(a^{2}+2)/3 = (3q+1){(3q+1)^{2}+2}/3=(3q+1) (9q^{2}+6q+1+2)/3= (3q+1) (9q^{2}+6q+3)/3= (3q+1) (3q^{2}+2q+1)

so a(a^{2}+1)/3 is an integer in this also.

*Finally, for a = 3q + 2, we see that*

a(a^{2 }+ 1)/3 = (3q + 2){(3q + 2)^{2}+ 2}/3 = (3q + 2)(9q^{2} + 12q +6)/3

=(3q + 2) (3q^{2} + 4q +2)

Consequently our result is established foe all cases. **(Proved)**