11.5. Can the integers $51 \sim 150$ be placed in a $10 \times 10$ table such that for any two adjacent cells (i.e., they share a common edge) with numbers $a$ and $b$, at least one of the equations $x^{2}-a x+b=0$ and $x^{2}-b x+a=0$ has two integer roots.
11.5. Not possible.

Assume there is a way to satisfy the requirement. Let $770$. This indicates that $1, a$ are its roots, and $b=1+a$.

If the equation $x^{2}-a x+b=0$ has two integer roots $x_{1}, x_{2}$, then $x_{1}+x_{2}=a, x_{1} x_{2}=b$;
If there are also $x_{1}, x_{2} \geqslant 2$, then
$$
b=x_{1}\left(a-x_{1}\right) \geqslant 2(a-2),
$$

This implies $b \geqslant 2 a-4>150$, a contradiction.
Therefore, 1 is a root of the equation,
$$
b=1 \times(a-1)=a-1 \text{. }
$$

The above discussion shows that for such a prime number $a$, there are only two numbers that can be adjacent to it, i.e.,
$$
b=a-1, b=a+1 \text{. }
$$

Thus, $a$ can only be located at the corners of the table. However, there are more than 4 prime numbers between 77 and 150 (such as $79, 83, 89, 97, 101$), and they cannot all be located at the corners. Contradiction.