# Question about estimating random symmetric sums modulo p

Let $$n > 0$$ be a positive integer (large) and $$p > 2$$ a fixed prime number. What is the probability that $$\sum_{ 1 \leq i < j \leq n} a_ia_j = 0 \mod p$$ where $$a_1, a_2, \dots a_n$$ are chosen uniformly from the set $$S = \{-1, 1\}$$. Does this sum equidistribute mod $$p$$ as $$n$$ goes to infinity? What would be the speed of equidistribution in terms of $$n$$? Is there any literature in this type of random sums? I would be surprised if not but I am unable to find anything related or similar to this.

One can also ask what is the probability of this sum being actually zero, but I also have no idea how to deal with it and thought that modulo a prime would be simpler.

The condition $$\sum_{ 1 \leq i < j \leq n} a_ia_j \equiv 0 \pmod p$$ is equivalent to $$\left(\sum_{ 1 \leq i\leq n} a_i\right)^2 \equiv n \pmod p.$$ So a necessary condition is that $$n$$ is a quadratic residue modulo $$p$$ (including the zero residue). If $$n$$ is divisible by $$p$$, then the above condition says that the sum of the $$a_i$$'s is divisible by $$p$$. Otherwise, the condition says that the sum of the $$a_i$$'s is congruent to one of the two square-roots of $$n$$ modulo $$p$$. Now it is easy to see that the sum of the $$a_i$$'s is equidistributed modulo $$p$$ (think about what happens when an $$a_i=1$$ is switched to $$a_i=-1$$), hence in the first case the probability is $$1/p+o(1)$$, in the second case it is $$2/p+o(1)$$, as $$n$$ tends to infinity.
In fact the probabilities can be calculated explicitly as a linear combination of $$n$$-th powers of $$p$$ complex numbers (which only depend on $$p$$), since the sum of the $$a_i$$'s modulo $$p$$ is determined by $$\#\{i:a_i=1\}$$ modulo $$p$$, and vice versa. Compare with this post, where the role of $$p$$ is played by $$4$$. It follows, in particular, that the $$o(1)$$ terms above decay exponentially fast. For a more complete reference, see Theorems 8.7.2 & 8.7.3 in Wagner: A first course in enumerative combinatorics (AMS, 2020).
• Dear @GHfromMO, thanks a lot for the answer and for the reference. Your answer depends on the lucky coincidence that $\sum a_i^2 = n$, what about the case where $S \neq {-1,1}$?. Also,do you know of a trick that can help with the last sums which includes a_i's and b_i's?, observe that these sums are not symmetric and even in the case $S = {-1,1}$ I do not know how to approach them. Aug 18 '21 at 23:42