Let $p > 2$ be a prime of bounded size. Suppose $A$ is a subset of $G = \mathbf{F}_p^n$ with only degenerate solutions to $$x + y = 2a,\\ x+z = 2b,\\ y + z = 2c,$$ where a solution is considered degenerate if any of $x,y,z$ are equal. Is there some $\theta = \theta(p) < p$ such that $|A| \leq \theta^p$?
My question (which arose in discussions with Freddie Manners) is motivated by trying to understand the limitations of the slice rank method. The above system has complexity 1 in the sense of additive combinatorics, so it is "controlled by Fourier analysis", and considered not much more difficult than the 3AP system, but it does not seem to yield to an obvious adaptation of the solution for ordinary capsets, as far as I can see.
In applications of the slice rank method, there are normally two steps: bound the slice rank above and bound it below. For this system, neither part is clear.
To bound the slice rank above, one expresses the configuration tensor $T$ in a clever way that exhibits low slice rank. In this case, it looks natural to write $$ \sum_{\text{solutions}} e_x \otimes \cdots \otimes e_c = \sum_{x,y,z,a,b,c\in G} \prod_{i=1}^n (1 - (x_i+y_i-2a_i)^{p-1}) (1 - (x_i+z_i-2b_i)^{p-1}) (1 - (y_i+z_i-2c_i)^{p-1}). $$ But as a polynomial this has total degree $3(p-1)n$, which is exactly typical for a polynomial in 6 variables, so no obvious bound for the slice rank comes out of this.
To bound the slice rank below, one restricts $T$ to $A^6$ and uses the diagonal-like structure. This part is not clear either, because the presence of partially degenerate solutions muddy the water. If $A$ has only degenerate solutions then $T$ is a sum of terms like $$ e_x \otimes e_x \otimes e_z \otimes e_x \otimes e_{(x+z)/2} \otimes e_{(x+z)/2},$$ and it's not clear why such a tensor should have large slice rank.
