The Question says prove that the square of every prime number greater than 3 yields a remainder of 1 when divided by 12.
My approach was that since every prime can be written as $6k\pm 1$ and when we square this expression we get in both cases $36k^2+12k+1$. Taking 12 common from the first two terms of the expression we get $12(3k^2+k)+1$.
So $12a+1$ where $a = (3k^2+k)$ yields a remainder 1 upon division with 12 right? Is this correct and if it is is there a more elegant proof?
Any help would be appreciated.