Can you generalize this?

$x^2-x(21+b)+21b-5=0$

For integral roots, $Discriminant$ should be a perfect square.

$(21+b)^2-4(21b-5)=k^2 ......k\in Z$

$(21-b)^2-k^2=-20$

$(21-b+k)(21-b-k)=-20$

The only possible factors are $-10,2$ and $-2,10$ for $-20$ so that $b$ takes integral values.

$21-b+k=10$ and $21-b-k=-2$

or

$21-b+k=-10$ and $21-b-k=2$

On solving,

$b=17$ and $b=25$ .

Difference= $25-17=8$

If u know a better method do post....

It is given that $(x-b)(x-21)=5\quad \Rightarrow b = x - \dfrac {5}{x-21}$ .

For $b$ to be integer, $\dfrac{5}{x-21}$ must be an integer and there are only $4$ cases:

When $x - 21 = \begin{cases} -5 & \Rightarrow x = 16 & \Rightarrow b = 16 + 1 = 17 \\ -1 & \Rightarrow x = 20 & \Rightarrow b = 20 + 5 = 25 \\ 1 & \Rightarrow x = 22 & \Rightarrow b = 22 - 5 = 17 \\ 5 & \Rightarrow x = 26 & \Rightarrow b = 26 - 1 = 25 \end{cases}$

$\Rightarrow b_{max} - b_{min} = 25 - 17 = \boxed{8}$

$\text{Since 5 is prime, either}(x-b)\text{or}(x-21)\text{has the value 5.}\\ \text{(since there are integral roots.)}\\ \text{If }x=26\text{ then b yields 25}\\ \text{ (since}(x-b)=1.)\\ \text{else, if }x=22\text{ then b=17}\\ \text{(Since} (x-b)=5 \text{ here.)}\\25-17=8.$