Given that $a,b,c,d \in Z, a\neq b\neq c \neq d$ are in arithmetic progression, and $$d = a^2 + b^2 + c^2$$ Find the general term of the Arithmetic Progression.
My work:
Select a = $\alpha - \beta$, b = $\alpha$, c = $\alpha + \beta$, d = $\alpha + 2\beta$.
Then we obtain:
$\alpha + 2\beta$ = 3$\alpha^2+2\beta^2$ or
$$\alpha(3\alpha - 1) = 2\beta(1-\beta)$$
I'm unsure how to proceed other than hit and trial.
$3\alpha^2-\alpha+2\beta^2-2\beta=0$, $36\alpha^2-12\alpha+6(4\beta^2-4\beta)=0$, $(6\alpha-1)^2+6(2\beta-1)^2=7$, so $\alpha$ must be zero, and $2\beta-1$ must be $\pm1$, so $\beta$ is zero or one. But $\beta=0$ contradicts $a\ne b$, so $\alpha=0$, $\beta=1$.