Re-arranging the given equation to one with respect to $a$: \begin{equation} a^2 -(2x^2-16x+21) a +(x^4-16x^3+81x^2-142x+68)=0 \end{equation} Because $$x^4-16x^3+81x^2-142x+68=(x^2-6x+4)(x^2-10x+17),$$ and $$(x^2-6x+4)+(x^2-10x+17) = (2x^2-16x+21)$$ The $1^{st}$ equation can be solved using factorization: $$a=x^2 - 6x + 4\quad\text{or}\quad a=x^2-10x+17$$ This implies that $$x^2 - 6x + (4-a) =0\quad\text{or}\quad x^2-10x+(17-a)=0 $$ Solving these two equations gives $$x_{1,2}=3\pm\sqrt{5+a}\quad\quad x_{3,4}=5\pm\sqrt{8+a}$$ In order for all $x$ to be integers, both $5+a$ and $8+a$ must be perfect squares. Let $$5+a = m^2\quad\text{and}\quad 8 + a = n^2$$ where $m$ and $n$ are two positive integers satisfying $n > m$. Hence $n^2-m^2 = 3 \implies (n+m)(n-m)=3$. The only solution is $n=2$ and $m=1$. It follows $$a=\boxed{-4}, x_{1,2,3,4}=\boxed{2, 4, 3, 7}$$