A.P numbers

Let $t$ be the difference. Therefore, $a+3t={ a }^{ 2 }+{(a+t)}^{2}+{(a+2t)}^{2}$ . Upon simplifying, $5{ t }^{ 2 }+t(6a-3)+3{ a }^{ 2 }-a=0$ , a quadratic in $t$ . Since $t$ is a real number, $D\ge 0$ . Thus, ${ (6a-3) }^{ 2 }\ge 20(3{ a }^{ 2 }-a)$ . Upon simplifying, $24{ a }^{ 2 }+16a-9\le 0$ . This gives an inequality for $a$ : $-\frac { 1 }{ 3 } -\frac { \sqrt { 70 } }{ 12 } <a<-\frac { 1 }{ 3 } +\frac { \sqrt { 70 } }{ 12 }$ . Since $a$ is an integer, $a=-1,0$ . If $a=0$ , $t=0,\frac { 3 }{ 5 }$ . Because $b,c,d$ are distinct integers, $a$ is not $0$ . Hence, $a=-1$ and $t=1$ . The integers are $-1,0,1,2$ . Thus the sum is $2$ .

If $k$ is the common difference, then it is required that $b+2k=3b^2+2k^2$ or $3b^2-b=2k-2k^2$ . Now $3b^2-b\geq 0$ so $k-k^2\geq 0$ and $k=0$ or $k=1$ . Since the integers are required to be distinct, we have $k=1,b=0$ . The progression is $-1,0,1,2$ and the answer is $\boxed{2}$ .