Consider an odd prime $p\equiv1 \pmod {16}$ and set $M=\frac{p-1}{2}$ for notational convenience. Then is there even a single prime $p$ of the above form for which the following congruence holds? $$\binom{M}{M/2}\binom{M}{M/4}\equiv \pm \binom{M}{3M/8}\binom{M}{7M/8}\pmod p ?$$
Here the symbol $\binom{n}{r}$ is a binomial coefficient and is defined as $\frac{n!}{r!(n-r)!}$
I am trying to prove something significantly more abstract but have managed to reduce it down to proving that the above statement can never hold. I have written some code to check this for primes of the above form less than $1000$ and found no counter examples. My attempts to find results on binomial coefficients $\pmod p$ have mostly involved Lucas' theorem which does not seem very applicable here. I would appreciate a solution, reference or possible strategies I might try.
Added later: The more general problem.
Let $$f_a(x)=x(x^4-1)(x^4+ax^2+1)$$ be a family of polynomials in $ \mathbb F_p[t]$ in the parameter $a\in \overline {\mathbb F}_p$.
Let $$\sum c_{\alpha}x^\alpha := f_a(x)^{\frac{p-1}{2}} $$ and note that the $c_\alpha$ are polynomials in $a$. Now make the $4 \times 4$ matrix whose $(i,j)$th entry is $c_{pi-j}$ for $1\leq i,j \leq 4$. Then I want to show that this matrix is non-invertible for values of $a$ other than $a=2,-2$. In other words, I want to show that the determinant $d(a)$ has irreducible factors other than $(a-2)$ and $(a+2)$.
What I have managed to prove is that $d(a)$ is a polynomial of degree $\frac{3(p-1)}{2}$ and also that $(a-2)$ and $(a+2)$ occur with the same multiplicity in the factorization of $d(a)$ in $\mathbb F_p[a]$. So for a contradiction, suppose $d(a)=\alpha(a^2-4)^{\frac{3(p-1)}{4}}$ for some $\alpha\in \mathbb F_p$. Then, the ratio of the leading term and the constant term must be 1. The leading term is given by the square of the right hand side in the top most equation and the constant term is given by the square of the left hand side. The other coefficients are very difficult to extract and it is lucky that I even managed to get the leading and constant terms. So I am hoping that if by some combinatorial argument I can rule out this congruence, then I will have proved that the determinant has to have an irreducible factor other than $(a-2)$ and $(a+2)$.