Parity congruence conjecture for finite Mordell--Tornheim sums

Let $r\in\N$, let $p$ be a prime, and let $\bfs\in\N^d$ satisfy $p>|\bfs|+1$, where $|\bfs|$ is the weight and $Z_{p^r}(\bfs)$ is the corresponding finite Mordell--Tornheim sum. Finite Mordell--Tornheim congruence conjecture. The following congruences should hold:

1. If $d=4$ and $|\bfs|$ is odd, then

$$Z_{p^r}(\bfs)\equiv p^{2r-2}Z_p(\bfs)\pmod{p^{2r-1}}.$$

2. If $d=4$ and $|\bfs|$ is even, then

$$Z_{p^r}(\bfs)\equiv p^{r-1}Z_p(\bfs)\pmod{p^r}.$$

3. If $d=5$ and $|\bfs|$ is even, then

$$Z_{p^r}(\bfs)\equiv p^{2r-2}Z_p(\bfs)\pmod{p^{2r-1}}.$$

In general, if $r\geq2$ and $d+|\bfs|$ is odd, then

$$Z_{p^r}(\bfs)\equiv0\pmod{p^{2r-2}};$$

if $r\geq2$ and $d+|\bfs|$ is even, then

$$Z_{p^r}(\bfs)\equiv0\pmod{p^{r-1}}.$$

The conjecture extends parity-dependent congruences for finite Mordell--Tornheim sums beyond the cases proved earlier in the paper. It is supported by extensive numerical evidence, while the stated general assertions remain open.