If distinct integers $a,b,c,d$ are consecutive terms of an arithmetic progression such that . . .

Hint:

Let $a,b,c,d$ be $A-3D, A-D, A+D, A+3D$

$$A+3D=(A-3D)^2+(A-D)^2+(A+D)^2=3A^2+11D^2-6AD$$

$$3A^2-A(6D+1)+11D^2-3D=0$$

The discriminant $$(6D+1)^2-12(11D^2-3D)=-96D^2+48D+1=1+6-96\left(D-\dfrac14\right)^2\le7$$ has to be perfect square