jmc

In an effort to factor this quartic polynomial, we try completing the square. If we square $x^2+p,$ then we get $$(x^2+p{)}^2=x^4+2p{x}^2+p^2,$$which gives us a term of $x^4.$ Thus, $$\begin{array}{rl}{x}^4-4x-1& =(x^2+p{)}^2-2p{x}^2-p^2-4x-1\\ & =(x^2+p{)}^2-(2p{x}^2+4x+p^2+1).\end{array}$$If we can choose a value of $p$ such that $2p{x}^2+4x+p^2+1$ is the square of a binomial, then we can factor the quartic using the difference-of-squares factorization.

The quadratic $2p{x}^2+4x+p^2+1$ is a perfect square if and only if its discriminant is 0, so $$4^2-4(2p)(p^2+1)=0.$$This simplifies to $p^3+p-2=0.$ We see that $p=1$ is a root.

Then for $p=1,$ we get $$\begin{array}{rl}{x}^4-4x-1& =(x^2+1{)}^2-(2x^2+4x+2)\\ & =(x^2+1)-2(x^2+2x+1)\\ & =(x^2+1)-[(x+1)\sqrt{2}{]}^2\\ & =(x^2+(x+1)\sqrt{2}+1)(x^2-(x+1)\sqrt{2}+1)\\ & =(x^2+x\sqrt{2}+\sqrt{2}+1)(x^2-x\sqrt{2}-\sqrt{2}+1).\end{array}$$The discriminant of the first quadratic factor is negative, so it has no real roots. The discriminant of the second quadratic factor is positive, so $a$ and $b$ are the roots of this quadratic.

Then by Vieta's formulas, $a+b=\sqrt{2}$ and $ab=-\sqrt{2}+1,$ so $ab+a+b=\boxed{1}.$