Beautiful Progression!

Let the terms be $a-3d,a-d,a+d,a+3d$

Since this is an increasing AP, it is obvious that the largest term will be equal to the sum of squares of the other three. If any other term is such, then the square of the largest number will automatically be greater than that term. An important observation is that all the terms cannot be negative since the largest one is equal to the sum of squares of the rest which makes it positive. And hence it is certain that the largest term will be equal to the sum of squares of the other three.

Given : $a+3d= (a-3d)^2 +(a-d)^2 +(a+d)^2$

Solving this quadratic equation in $a$ ,

We get $d=0, \frac{1}{2}$

However $d \ne 0$ because then all the terms would be equal and there would not be an increasing AP.

$\therefore d= \frac{1}{2}$

Substituting $d= \frac{1}{2}$ , we get $a= \frac{1}{2} or \frac{5}{6}$

But, $a \ne \frac{5}{6}$ because then $a-d$ would equal $\frac{1}{3}$ which is not an integer.

$\therefore a= \frac{1}{2}$

Substituting values of $a$ and $d$ ,

The four terms are $-1,0,1,2$

$\therefore -1 + 1 + (-1+0+1+2) + (-1)(0)(1)(2) = \large \boxed{2}$

I tried the problem with the exact method but calculations seemed to loop.

This is not a solution but a hit & trial way of doing this solution. An AP for these conditions can be -1, 0, 1,2.