A Diophantine equation is almost-satisfiable iff it is satisfiable over the ring $\widehat{\mathbb Z}$, the profinite completion of $\mathbb Z$ (also called by some the Prufer ring), by a standard compactness argument (the solutions over each $\mathbb Z/n\mathbb Z$ form an inverse system of finite sets whose inverse limit is the set of solutions over $\widehat{\mathbb Z}$, and hence if the set of soltuions is non-empty for each $n$, then the inverse limit is non-empty). From the direct product decomposition $\widehat{\mathbb Z}=\prod_p \mathbb Z_p$, this in turn reduces things to solvability over the $p$-adic integers $\mathbb Z_p$ for all $p$. This question is solved in the paper J. Ax, Solving diophantine problems modulo every prime. Ann. Math. 85, 161–183 (1967) on pages 170,171. So the problem is decidable.

A Diophantine equation is almost-satisfiable iff it is satisfiable over the ring $\widehat{\mathbb Z}$, the profinite completion of $\mathbb Z$ (also called by some the Prüfer ring), by a standard compactness argument (the solutions over each $\mathbb Z/n\mathbb Z$ form an inverse system of finite sets whose inverse limit is the set of solutions over $\widehat{\mathbb Z}$, and hence if the set of solutions is non-empty for each $n$, then the inverse limit is non-empty). From the direct product decomposition $\widehat{\mathbb Z}=\prod_p \mathbb Z_p$, this in turn reduces things to solvability over the $p$-adic integers $\mathbb Z_p$ for all $p$. This question is solved in the paper J. Ax, Solving diophantine problems modulo every prime. Ann. Math. 85, 161–183 (1967) on pages 170,171. So the problem is decidable.