What is the value of $1^2 + 2^2 + 3^2 + \cdots + (p-1)^2\pmod{p}$? What is the value of $1^2 + 2^2 + 3^2 + \cdots + (p-1)^2\pmod{p}$?
Let's try a several primes greater than 3... 
If $p=5$, then we have $1^2 + 2^2 + 3^2 + 4^2 = 30$, so that $30\pmod{5} = 0$
If $p=7$, then we have $1^2 + 2^2 + 3^2 + 4^2 + 5^2 + 6^2= 91$, so that $91\pmod{7} = 0$
If $p=11$, then we have $1^2 + 2^2 + 3^2 + 4^2 + 5^2 + 6^2 + 7^2 + 8^2 + 9^2 + 10^2= 385$, so that $385\pmod{11} = 0$
So it seems like we will always get 0. So for primes greater than 3, we have the sum equal to a multiple of $p$.
I know that $1^2 + 2^2 + 3^2 + \cdots + (p-1)^2 = \frac16p(p-1)(2p-1)$.Well if $p>3$, then $p-1=2a$ for some integer $a$, so that $2p-1 = 4a +1$. Then we have 
$$ 1^2 + 2^2 + 3^2 + \cdots + (p-1)^2 = \frac16p(p-1)(2p-1) = \frac16 p(2a)(4a+1)$$
If I could show that $(2a)(4a+1)$ is a multiple of 6, I'll be done, but I am having trouble showing that. Perhaps someone could provide a much needed tip, but don't figure this out for me, please. 

Assuming the formula for sum of squares (which has to be an integer) we see that  the 6 in the denominator has to divide the numerator: Now 6 is coprime to $p$ (for $p >3$). Now you can figure it out yourself.
