Given function

$f(x)=(x-p)^2-q, \qquad q>0$

We are told that $a_1,a_2,a_3,a_4$ are in an arithmetic progression with mean $p$ and positive common difference.

Write the A.P. in symmetric form

Since four numbers are in A.P. with mean $p$, they must be symmetric about $p$:

$a_1=p-3d,\quad a_2=p-d,\quad a_3=p+d,\quad a_4=p+3d$

where $d>0$.

Use the condition $|f(a_i)|=500$

Now,

$f(a_1)=f(p-3d)=( -3d)^2-q=9d^2-q$ $f(a_2)=f(p-d)=d^2-q$ $f(a_3)=f(p+d)=d^2-q$ $f(a_4)=f(p+3d)=9d^2-q$

Since $|f(a_i)|=500$ for all $i$, we get

$|d^2-q|=500$ and $|9d^2-q|=500$

Solve the two absolute value equations

Let $A=d^2-q, \qquad B=9d^2-q$ with $|A|=500, \qquad |B|=500$

So $A,B\in\{500,-500\}$.

Also, $B-A=(9d^2-q)-(d^2-q)=8d^2$

Since $d>0$, we have $8d^2>0$.

Check possibilities:

• If $A=500, B=500$, then $B-A=0$ impossible.
• If $A=-500, B=-500$, then $B-A=0$ impossible.
• If $A=500, B=-500$, then $B-A=-1000$ impossible.
• If $A=-500, B=500$, then $B-A=1000=8d^2$ $d^2=125$

Then from $A=d^2-q=-500$, $125-q=-500$ $q=625$

Find the roots of $f(x)=0$

$f(x)=0 \implies (x-p)^2-q=0$ $(x-p)^2=q$ $x=p\pm \sqrt{q}$

Hence the absolute difference between the roots is

$|(p+\sqrt q)-(p-\sqrt q)|=2\sqrt q$

With $q=625$,

$2\sqrt{625}=2\cdot 25=50$

Final answer

$\boxed{50}$