The problem states that if a is odd, then prove 12 | a^2 + (a+2)^2 + (a+4)^2 + 1

Here is what I did but it seems more by induction than the division algorithm. If someone could let me know if I am on the right track.

Base case a=1 .... 12 |36 done.

So, if k is in G then prove (2k-1) is in G

12 | (2k-1)^2 + ((2k-1)+2)^2 + ((2k-1)+4)^2 +1 - I simply multiplied this whole thing out and got 12 | 12k^2 + 12k + 12

Then if a|b we know a|bc, so 12|12 or 12|-12 if we want to use if a|b and a|c then a|b-c

Thus k+1 is in G.

I guess I just thought I would be looking for k(k+1)(k+2) as we did in the example in class.