Finding the sum of all products of pairs of distinct primitive roots mod 83 I'm currently studying Number Theory and I've stumbled upon a question where I need to: Find the sum of all products of pairs of distinct primitive roots mod 83.
Solving attempt: I've tried to find all the primitive roots mod 83 but then I realized that there are probably many of them and the calculations are getting heavy on high powers. I guess there might be a simpler approach then just finding all the primitive roots and summing all the products of distinct primitive roots. I do know that the product of all primitive roots (mod p) is 1 mod p but I don't see how it helps me.
Any help would be appreciated.
 A: Let $g$ be a primitive root of $83$. Then all the primitive roots are $g^k$ where $k$ is relatively prime to $82$,  so all $g^k$ with odd $k$ from $k=1$ to $k=81$, with the exception of $k=41$. These are all the quadratic non-residues of $83$ except $-1$. 
Let $S$ be the sum of the primitive roots. Then the sum of the quadratic non-residues of $83$ is congruent to $S-1$ modulo $p$. This is $0$ modulo $83$, so $S\equiv 1\pmod{83}$.
The sum of the squares of the quadratic non-residues is congruent to $0$ modulo $83$. This is because $83$ is of the form $4k-1$, so $x$ is a QR if and only if $-x$ is an NR. So the sum $T$ of the squares of the primitive roots is congruent to $-1$.
Our sum of products is $\frac{1}{2}(S^2-T)$, which is congruent to $1$ modulo $p$. 
A: Given any polynomial $f(x)=a_nx^n+a_{n-1}x^{n-1}+\dots+a_1x+a_0\in\mathbb C[x]$ with complex roots $p_1,\dots,p_n$ (counted with multiplicity), we have
$$
\sum_{i\neq j}p_ip_j=\frac{a_{n-2}}{a_n}
$$
following from Vieta's formulas. Consider the cyclotomic polynomial $$\Phi_{83}(x)=\prod_{\zeta\text{ primitive $83^\text{th}$ root}}(x-\zeta).$$
Since $83$ is prime, there is a formula you can use to calculate $\Phi_{83}$. Can you finish it from here?
