Let $A$,$B$ two integer, symmetric, $n\times n$ matrix with even diagonal. We suppose also that $A$ has odd determinant (so $n$ is even).
Even discarding the hypothesis of even diagonal for $A$ it's not difficult to show that:
$$\det(A)\equiv\det(A+2B)\pmod4$$
But I suspect that if $A$ is even diagonal too then we have the stronger equivalence: $$\det(A)\equiv\det(A+2B)\pmod8$$
How could I prove it?