A NP-complete problem with difficult NP membership proof

Question

A proof of $\mathsf{NP}$-completeness for a decision problem requires two things:

• $\mathsf{NP}$ membership;
• $\mathsf{NP}$-hardness.

It is often the case that the $\mathsf{NP}$-hardness part is the most difficult to prove, and it is not rare that some papers completely omit the $\mathsf{NP}$ membership part.

What I would like to ask is: is there a $\mathsf{NP}$-complete problem such that its $\mathsf{NP}$ membership via certificate existence is at least as difficult to prove than its $\mathsf{NP}$-hardness. For example, a primality certificate is not quite intuitive to find (but the primality decision problem is in $\mathsf{P}$, so it is not $\mathsf{NP}$-complete unless $\mathsf{P} = \mathsf{NP}$).

Answer 1

Integer Linear Programming is such a problem.

• Input: two matrices $A\in \mathcal{M}_n(\mathbb{Z})$ and $B\in \mathcal{M}_{n,1}(\mathbb{Z})$.
• Question: is there a matrix $X\in \mathcal{M}_{n,1}(\mathbb{Z})$ such that $AX\leqslant B$? (the inequality being component by component).

An obvious certificate would be such a matrix $X$, but it is very hard to prove that there exists such an $X$ with size polynomial in the size of the input (indeed, the coefficient of $X$ could have a number of digits exponential in the number of digits of the coefficients of $A$ or $B$).