If you have $n$ identically distributed Bernoulli trials whose sum is binomially distributed random variable, does it then follow that the $n$ Bernoulli trials are independent?
That's false even for $n=3$.
Denote by $B(p)$ the Bernoulli distribution which has probability $p$ of being 1 and $(1-p)$ to be 0.
Define the three Bernoulli variables $(X_1, X_2, X_3)$ by $ X_1 \sim B(0.5) \\ X_2|(X_1=1) \sim B(0.25);~ X_2|(X_1=0) \sim B(0.75) \\ X_3|(X_1+X_2=2) \sim B(1);~ X_3|(X_1+X_2=0) \sim B(0);~ X_3|(X_1+X_2=1) \sim B(0.5) $
Quick calculation shows they're all $B(0.5)$ distributed (basically by symmetry) and that the sum has the same distribution as the sum of three i.i.d. trials.
The answer is no for large enough $n$, because the independence of the $n$ Bernoulli random variables (r.v.'s) is given by about $2^n$ equations, whereas to describe the individual distributions of the Bernoulli r.v.'s and their sum one needs only $O(n)$ equations.