To prove that it is easy to verify, take any two numbers from $2, 5, 13$ as $a, b$, then $ab-1$ must be a perfect square. Therefore, the selection of $a, b$ must be one as $d$, and the other chosen from $2, 5, 13$. We will use proof by contradiction to show that there must be a selection such that $ab-1$ is not a perfect square. If $ab-1$ are all perfect squares, then we have
$$2d-1=x^2, \quad 5d-1=y^2, \quad 13d-1=z^2$$
If equation (1) holds, it indicates that $d$ has a close relationship with the sequence of perfect squares, thus it has strong restrictions on the remainders of certain moduli. We will point out that these restrictions are contradictory. This is the idea to solve this problem.

From the first equation of (1), we know that $x$ is an odd number, and $x^2$ is a number of the form $8m+1$, so
$$d=4k+1$$

From this and the second and third equations of (1), we deduce that $y, z$ are both even, set as
$$y=2u, \quad z=2v.$$

Using the above two equations, by subtracting the second and third equations of (1), we get
$$v^2-u^2=2d=8k+2$$

But $v^2-u^2$ cannot equal $8k+2$, which is a contradiction.