Given an odd integer $a$ , establish that $a^2+(a+2)^2+(a+4)^2+1$ is divisible by $12$? Given an odd integer $a$ , establish that $a^2+(a+2)^2+(a+4)^2+1$ is divisible by $12$?
So far I have:
$a^2+(a+2)^2+(a+4)^2+1$ 
$=a^2+a^2+4a+4+a^2+8a+16+1 $
$=3a^2+12a+21$ 
$=3(a^2+4a+7) $
where do I go from here.. the solution I have is divisible by $3$ not $12$...

Write $a = 2t+1$. Then $a^2+(a+2)^2+(a+4)^2+1=12 (t^2+3 t+3)$.
It may be simpler to write $b=a+2$. Then $a^2+(a+2)^2+(a+4)^2+1=(b-2)^2+b^2+(b+2)^2+1=3b^2+9$. Now $a$ odd implies $b$ odd, and so write $b=2u+1$. Then $3b^2+9=12 (u^2+u+1)$.
