<s>[INST] 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 $12$...
 [/INST] $$a^2+(a+2)^2+(a+4)^2+1=3a^2+12a+21\equiv3(a^2-1)\pmod{12}$$
Now, if $a$ is odd $a\pm1$ are even $\implies a^2-1$ will be divisible by $4$
In fact, $\displaystyle(2b+1)^2=4b^2+4b+1=8\frac{b(b+1)}2+1\equiv1\pmod8$
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